Application-Oriented Examples - Analog and Digital Signal Processing

In this section, we analyze several circuits that are useful in understanding the concepts of signal processing and discuss some of their performance characteristics based on the results of the analysis.

4.7.1 An RC Lowpass Filter

Consider the relaxed RC lowpass filter sketched in the following review panel. Writing a node equation (based on Kirchhoﬀ’s current law) at the node labeled A, we get the result

y′(t) + 1 τy(t) =

τx(t) (4.23)

where τ = RC defines the time constant of the circuit. The characteristic equation s + 1

τ = 0 yields the natural response yN(t) = Ke−t/τ. To find the step response, we let x(t) = u(t) = 1, t ≥ 0. So the forced response, yF(t) = B, is a constant. Then, with yF′ (t) = 0, we obtain

y_F′ (t) +1

τyF(t) = 1

τ = 0 + B or B = 1

Thus, y(t) = yF(t) + yN(t) = 1 + Ke−t/τ. With y(0) = 0, we get 0 = 1 + K and

s(t) = y(t) = (1− e−t/τ_)u(t) _{(step response)} _(4.24)

The impulse response h(t) equals the derivative of the step response. Thus, h(t) = s′(t) = 1

τe

−t/τ_u(t) _{(impulse response)} _(4.25)

REVIEW PANEL 4.19

Unit Step Response and Unit Impulse Response of an RC Lowpass Filter The output is the capacitor voltage. The time constant is τ = RC.

t /τ - e 1− t 1 s(t) t (1) 1 t R C + + − − A τ 1 t /τ - τ 1_e h(t) t R C + + − −

Step response: s(t) = (1 − e−t/τ_)u(t) _{Impulse response: h(t) = s}′_{(t) =} 1

τe−t/τu(t)

Performance Measures

The time-domain performance of systems is often measured in terms of their impulse response and/or step response. For an exponential signal Ae−t/τ_{, the smaller the time constant τ, the faster is the decay. For} first-order systems, the time constant τ is a useful measure of the speed of the response, as illustrated in Figure 4.1.

The smaller the time constant, the faster the system responds, and the more the output resembles (matches) the applied input. An exponential decays to less than 1% of its peak value in about 5τ. As a result, the step response is also within 1% of its final value in about 5τ. This forms the basis for the observation that it takes about 5τ to reach steady state. For higher-order systems, the rate of decay and the

4.7 Application-Oriented Examples 87 − t /τ e 0.5A 0.9A 0.1A Small τ Large τ Large τ Small τ A t Exponential A _A t Delay t A

Step response Step response

Rise time

Figure 4.1 Time-domain performance measures

time to reach steady state depends on the largest time constant τmax (corresponding to the slowest decay) associated with the exponential terms in its impulse response. A smaller τmaximplies a faster response and a shorter time to reach steady state. The speed of response is also measured by the rise time, which is often defined as the time it takes for the step response to rise from 10% to 90% of its final value. Another useful measure of system performance is the delay time, which is often defined as the time it takes for the step response to reach 50% of its final value. These measures are also illustrated in Figure 4.1. Another measure is the settling time, defined as the time it takes for the step response to settle to within a small fraction (typically, 5%) of its final value.

REVIEW PANEL 4.20 Time Domain Measures of System Performance

Time constant τ The smaller the τ, the more the output resembles the input. Speed of response Depends on the largest time constant τmax in h(t).

Steady state Steady state is reached in about 5τmax.

Rise time Time for step response to rise from 10% to 90% of its final value. Delay time Time for step response to reach 50% of its final value.

5% settling time Time for step response to settle to within 5% of its final value. EXAMPLE 4.12 (Some Analog Filters)

Find the step response and impulse response of the circuits shown in Figure E4.12.

3 1 1F − + − x(t) y(t)

Second-order Bessel filter H 1Ω + y(t) x(t) _1F + − 1H + − y(t) x(t) _1F 1Ω 1F 1Ω + − 2H + −

Second-order Butterworth filter Third-order Butterworth filter Ω

Figure E4.12. Circuits for Example 4.12.

(a) (A Second-Order Bessel Filter)

The governing diﬀerential equation is y′′_{(t) + 3y}′_{(t) + 3y(t) = 3x(t).} The characteristic equation is s2_{+ 3s + 3 = 0. Its roots are s}

1,2 = −3₂± j

√ 3

2 .

The natural response is thus yN(t) = e−3t/2[K1cos( √

3t/2) + K2sin( √

3t/2)]. For the step response, x(t) = 1, t ≥ 0, and we assume yF(t) = B.

Then y′′

88 Chapter 4 Analog Systems The total response is y(t) = 1 + e−3t/2_[K

1cos(

√

3t/2) + K2sin( √

3t/2)]. Using zero initial conditions, we get y(0) = 0 = 1 + K1 and y′(0) = 0 =

√ 3

2 K2−32K1.

Thus, K1= −1, K2= −√3, and y(t) = u(t) − e−3t/2[cos(√3t/2) +√3 sin(√3t/2)]u(t). The impulse response h(t) equals the derivative of the step response.

Thus, h(t) = y′_{(t) = 2}√_3e−3t/2_sin(√_3t/2)u(t).

The impulse response decays to nearly zero in about 3 s (five time constants). (b) (A Second-Order Butterworth Filter)

The governing diﬀerential equation is y′′_{(t) +}√_2y′_{(t) + y(t) = x(t).} The characteristic equation is s2₊√_{2s + 1 = 0. Its roots are s}

1,2= −√1₂± j√1₂.

The natural response is thus yN(t) = e−t/ √

2_[K

1cos(t/√2) + K2sin(t/√2)]. For the step response, x(t) = 1, t ≥ 0, and we assume yF(t) = B.

Then y′′ F(t) +

√ 2y′

F(t) + yF(t) = B = 1. The total response is y(t) = 1 + e−t/√2_[K

1cos(t/√2) + K2sin(t/√2)].

Using zero initial conditions, we get y(0) = 0 = 1 + K1 and y′(0) = 0 = −√1₂(K2− K1). Thus, K1= K2= −1 and y(t) = u(t) − e−t/

√

2_[cos(t/√_{2) + sin(t/}√_2)]u(t). The impulse response h(t) equals the derivative of the step response. Thus, h(t) = y′_{(t) =}√_2e−t/√2_sin(t/√_2)u(t).

The impulse response decays to nearly zero in about 7 s (five time constants). (c) (A Third-Order Butterworth Filter)

The governing diﬀerential equation is y′′′_{(t) + 2y}′′_{(t) + 2y}′_{(t) + y(t) =}1 2x(t). The characteristic equation is s3_{+ 2s}2_{+ 2s + 1 = 0. Its roots are s}

1,2= −1₂± j

√ 3

2 and s3= −1. The natural response is thus yN(t) = e−t/2[K1cos(

√

3t/2) + K2sin( √

3t/2)] + K3e−t. For the step response, x(t) = 1, t ≥ 0, and we assume yF(t) = B.

Then y′′′

F(t) + 2yF′′(t) + 2yF′ (t) + yF(t) = 0 + 0 + B = 12. The total response is y(t) = 1

2+ e−3t/2[K1cos(

√_{3t/2) + K}

2sin(√3t/2)] + K3e−t. Using zero initial conditions, we get

y(0) = 0 = 1₂+ K1+ K3 y′(0) = 0 = −12K1+ √ 3 2 K2− K3 y′′(0) = 0 = −12K1− √ 3 2 K2+ K3

Thus, K1= 0, K2= −√1₃, K3= 1₂, and y(t) =₂1u(t)−√1₃e−t/2sin(√3t/2)u(t) −1₂e−tu(t). The impulse response h(t) equals the derivative of the step response.

Thus, h(t) = e−t/2_[ 1 2√3sin( √ 3t/2) −1 2cos( √ 3t/2)]u(t) − 1 2e−tu(t).

Chapter 4 Problems 89

CHAPTER 4 PROBLEMS

DRILL AND REINFORCEMENT

4.1 (Operators) Which of the following describe linear operators? (a) O{ } = 4{ } (b) O{ } = 4{ } + 3 (c) y(t) =$t

−∞x(t) dt (d) O{ } = sin{ } (e) y(t) = x(4t) (f) y(t) = 4d x(t)

dt + 3x(t)

4.2 (System Classification) In each of the following systems, x(t) is the input and y(t) is the output. Classify each system in terms of linearity, time invariance, memory, and causality.

(a) y′′_{(t) + 3y}′_{(t) = 2x}′_{(t) + x(t)} _{(b) y}′′_{(t) + 3y(t)y}′_{(t) = 2x}′_{(t) + x(t)} (c) y′′_{(t) + 3tx(t)y}′_{(t) = 2x}′_(t) _{(d) y}′′_{(t) + 3y}′_{(t) = 2x}2_{(t) + x(t + 2)} (e) y(t) + 3 = x2_{(t) + 2x(t)} _{(f) y(t) = 2x(t + 1) + 5}

(g) y′′_{(t) + e}−t_y′_{(t) = |x}′_{(t − 1)|} _{(h) y(t) = x}2_{(t) + 2x(t + 1)} (i) y′′_{(t) + cos(2t)y}′_{(t) = x}′_{(t + 1)} _{(j) y(t) + t}$t

−∞y(t) dt = 2x(t) (k) y′_{(t) +}$t

0y(t) dt =|x′(t)| − x(t) (l) y′′(t) + t $t+1

0 y(t) dt = x′(t) + 2

4.3 (Classification) Classify the following systems in terms of their linearity, time invariance, causality, and memory.

(a) The modulation system y(t) = x(t)cos(2πf0t). (b) The modulation system y(t) = [A + x(t)]cos(2πf0t).

(e) The sampling system y(t) = x(t) ∞ % k=−∞

δ(t− kts).

4.4 (Forced Response) Evaluate the forced response of the following systems. (a) y′_{(t) + 2y(t) = u(t)} _{(b) y}′_{(t) + 2y(t) = cos(t)u(t)}

(c) y′_{(t) + 2y(t) = e}−t_u(t) _{(d) y}′_{(t) + 2y(t) = e}−2t_u(t) (e) y′_{(t) + 2y(t) = tu(t)} _{(f) y}′_{(t) + 2y(t) = te}−2t_u(t)

4.5 (Forced Response) Evaluate the forced response of the following systems. (a) y′′_{(t) + 5y}′_{(t) + 6y(t) = 3u(t)} _{(b) y}′′_{(t) + 5y}′_{(t) + 6y(t) = 6e}−t_u(t) (c) y′′_{(t) + 5y}′_{(t) + 6y(t) = 5 cos(t)u(t) (d) y}′′_{(t) + 5y}′_{(t) + 6y(t) = 2e}−2t_u(t)

(e) y′′_{(t) + 5y}′_{(t) + 6y(t) = 2tu(t)} _{(f) y}′′_{(t) + 5y}′_{(t) + 6y(t) = (6e}−t_{+ 2e}−2t_)u(t)

4.6 (Steady-State Response) The forced response of a system to sinusoidal inputs is termed the steady- state response. Evaluate the steady-state response of the following systems.

(a) y′_{(t) + 5y(t) = 2u(t)} _{(b) y}′_{(t) + y(t) = cos(t)u(t)} (c) y′_{(t) + 3y(t) = sin(t)u(t)} _{(d) y}′_{(t) + 4y(t) = cos(t) + sin(2t)} (e) y′′_{(t) + 5y}′_{(t) + 6y(t) = cos(3t)u(t)} _{(f) y}′′_{(t) + 4y}′_{(t) + 4y(t) = cos(2t)u(t)}

90 Chapter 4 Analog Systems 4.7 (Zero-State Response) Evaluate the zero-state response of the following systems.

(a) y′_{(t) + 2y(t) = u(t)} _{(b) y}′_{(t) + y(t) = cos(t)u(t)} (c) y′_{(t) + y(t) = r(t)} _{(d) y}′_{(t) + 3y(t) = e}−t_u(t) (e) y′_{(t) + 2y(t) = e}−2t_u(t) _{(f) y}′_{(t) + 2y(t) = e}−2t_cos(t)u(t)

4.8 (Zero-State Response) Evaluate the zero-state response of the following systems. (a) y′′_{(t) + 5y}′_{(t) + 6y(t) = 6u(t)} _{(b) y}′′_{(t) + 4y}′_{(t) + 3y(t) = 2e}−2t_u(t) (c) y′′_{(t) + 2y}′_{(t) + 2y(t) = 2e}−t_u(t) _{(d) y}′′_{(t) + 4y}′_{(t) + 5y(t) = cos(t)u(t)}

(e) y′′_{(t) + 4y}′_{(t) + 3y(t) = r(t)} _{(f) y}′′_{(t) + 5y}′_{(t) + 4y(t) = (2e}−t_{+ 2e}−3t_)u(t)

4.9 (System Response) Evaluate the natural, forced, zero-state, zero-input, and total response of the following systems.

(a) y′_{(t) + 5y(t) = u(t)} _{y(0) = 2} (b) y′_{(t) + 3y(t) = 2e}−2t_u(t) _{y(0) = 1} (c) y′_{(t) + 4y(t) = 8tu(t)} _{y(0) = 2} (d) y′_{(t) + 2y(t) = 2 cos(2t)u(t)} _{y(0) = 4} (e) y′_{(t) + 2y(t) = 2e}−2t_u(t) _{y(0) = 6} (f) y′_{(t) + 2y(t) = 2e}−2t_{cos(t)u(t) y(0) = 8}

4.10 (System Response) Evaluate the response y(t) of the following systems. (a) y′_{(t) + y(t) = 2x}′_{(t) + x(t)} _{x(t) = 4e}−2t_{u(t) y(0) = 2} (b) y′_{(t) + 3y(t) = 3x}′′_(t) _{x(t) = 4e}−2t_{u(t) y(0) = 0}

(d) y′_{(t) + 2y(t) = x(t) + 2x(t − 1)} _{x(t) = 4u(t)} _{y(0) = 0} (e) y′_{(t) + 2y(t) = x}′_{(t) − 2x(t − 1)} _{x(t) = 2e}−t_u(t) _{y(0) = 0} (f) y′_{(t) + 2y(t) = x}′′_{(t) − 2x}′_{(t − 1) + x(t − 2)} _{x(t) = 2e}−t_u(t) _{y(0) = 4}

4.11 (System Response) For each of the following, evaluate the natural, forced, zero-state, zero-input, and total response. Assume y′_{(0) = 1 and all other initial conditions zero.}

(a) y′′_{(t) + 5y}′_{(t) + 6y(t) = 6u(t)} _{y(0) = 0 y}′_{(0) = 1} (b) y′′_{(t) + 5y}′_{(t) + 6y(t) = 2e}−t_u(t) _{y(0) = 0 y}′_{(0) = 1} (c) y′′_{(t) + 4y}′_{(t) + 3y(t) = 36tu(t)} _{y(0) = 0 y}′_{(0) = 1} (d) y′′_{(t) + 4y}′_{(t) + 4y(t) = 2e}−2t_u(t) _{y(0) = 0 y}′_{(0) = 1} (e) y′′_{(t) + 4y}′_{(t) + 4y(t) = 8 cos(2t)u(t)} _{y(0) = 0 y}′_{(0) = 1}

(f) [(s + 1)2_{(s + 2)]y(t) = e}−2t_u(t) _{y(0) = 0 y}′_{(0) = 1 y}′′_{(0) = 0} 4.12 (System Response) Evaluate the response y(t) of the following systems.

(a) y′′_{(t) + 3y}′_{(t) + 2y(t) = 2x}′_{(t) + x(t)} _{x(t) = 4u(t)} _{y(0) = 2 y}′_{(0) = 1} (b) y′′_{(t) + 4y}′_{(t) + 3y(t) = 3x}′′_(t) _{x(t) = 4e}−2t_{u(t) y(0) = 0 y}′_{(0) = 0} (c) y′′_{(t) + 4y}′_{(t) + 4y(t) = x}′_{(t) − x(t)} _{x(t) = 4u(t)} _{y(0) = 6 y}′_{(0) = −3} (d) y′′_{(t) + 2y}′_{(t) + 2y(t) = x(t) + 2x(t − 1)} _{x(t) = 4u(t)} _{y(0) = 0 y}′_{(0) = 0} (e) y′′_{(t) + 5y}′_{(t) + 6y(t) = x}′_{(t) − 2x(t − 1)} _{x(t) = 2e}−t_u(t) _{y(0) = 0 y}′_{(0) = 0} (f) y′′_{(t) + 5y}′_{(t) + 4y(t) = x}′′_{(t) − 2x}′_{(t − 1)} _{x(t) = 3e}−t_u(t) _{y(0) = 4 y}′_{(0) = −4} 4.13 (Impulse Response) Find the impulse response of the following systems.

(a) y′_{(t) + 3y(t) = x(t)} _{(b) y}′_{(t) + 4y(t) = 2x(t)} (c) y′_{(t) + 2y(t) = x}′_{(t) − 2x(t) (d) y}′_{(t) + y(t) = x}′_{(t) − x(t)}

Chapter 4 Problems 91 4.14 (Impulse Response) Find the impulse response of the following systems.

(a) y′′_{(t) + 5y}′_{(t) + 4y(t) = x(t)} _{(b) y}′′_{(t) + 4y}′_{(t) + 4y(t) = 2x(t)} (c) y′′_{(t) + 4y}′_{(t) + 3y(t) = 2x}′_{(t) − x(t) (d) y}′′_{(t) + 2y}′_{(t) + y(t) = x}′′_{(t) + x}′_(t) 4.15 (Stability) Which of the following systems are stable, and why?

(a) y′_{(t) + 4y(t) = x(t)} _{(b) y}′_{(t) − 4y(t) = 3x(t)}

(c) y′_{(t) + 4y(t) = x}′_{(t) + 3x(t)} _{(d) y}′′_{(t) + 5y}′_{(t) + 4y(t) = 6x(t)} (e) y′′_{(t) + 4y(t) = 2x}′_{(t) − x(t) (f) y}′′_{(t) + 5y}′_{(t) + 6y(t) = x}′′_(t) (g) y′′_{(t) − 5y}′_{(t) + 4y(t) = x(t)} _{(h) y}′′_{(t) + 2y}′_{(t) − 3y(t) = 2x}′_(t)

4.16 (Impulse Response) The voltage input to a series RC circuit with a time constant τ is 1

αe−t/αu(t). (a) Find the analytical form for the capacitor voltage.

(b) Show that as α → 0, we obtain the impulse response h(t) of the RC circuit.

4.17 (System Response) The step response of an LTI system is given by s(t) = (1 − e−t_)u(t). (a) Establish its impulse response h(t) and sketch both s(t) and h(t).

(b) Evaluate and sketch the response y(t) to the input x(t) = rect(t − 0.5).

REVIEW AND EXPLORATION

4.18 (System Classification) Explain why the system y′′_{(t) + y}′_{(t) = t sin(t) is time invariant, whereas} the system y′′_{(t) + y}′_{(t) = tx(t), where x(t) = sin(t), is time varying.}

4.19 (System Classification) Investigate the linearity, time invariance, memory, causality, and stability of the following operations.

(a) y(t) = y(0) +! t 0 x(λ) dλ (b) y(t) = ! t 0 x(λ) dλ, t > 0 (c) y(t) =! t+1 t−1 x(λ) dλ (d) y(t) = ! t+α t x(λ + 2) dλ (e) y(t) =! t−α t x(λ− 2) dλ (f) y(t) =! t+α t−1 x(λ + 1) dλ

4.20 (Classification) Check the following for linearity, time invariance, memory, causality, and stability. (a) The time-scaling system y(t) = x(2t)

(b) The folding system y(t) = x(−t) (c) The time-scaling system y(t) = x(0.5t) (d) The sign-inversion system y(t) = sgn[x(t)]

(e) The rectifying system y(t) = |x(t)|

4.21 (Classification) Consider the two systems (1) y(t) = x(αt) and (2) y(t) = x(t + α). (a) For what values of α is each system linear?

(b) For what values of α is each system causal?

92 Chapter 4 Analog Systems 4.22 (System Response) Consider the relaxed system y′_{(t) + y(t) = x(t).}

(a) The input is x(t) = u(t). What is the response?

(b) Use the result of part (a) (and superposition) to find the response of this system to the input x1(t) shown in Figure P4.22.

(d) Use the result of part (c) (and superposition) to find the response of this system to the input x2(t) shown in Figure P4.22.

(e) How are the results of parts (a) and (b) related to the results of parts (c) and (d)?

1 x (t) 2 x (t) −4 4 1 2 t −4 4 t 2 1

Figure P4.22 Input signals for Problem 4.22

4.23 (System Response) Consider the relaxed system y′_{(t) +}1

τy(t) = x(t). (a) What is the response of this system to the unit step x(t) = u(t)? (b) What is the response of this system to the unit impulse x(t) = δ(t)?

(c) What is the response of this system to the rectangular pulse x(t) = u(t) − u(t − α)? Under what conditions for τ and α will the response resemble (be a good approximation to) the input? 4.24 (System Response) It is known that the response of the system y′_{(t) + αy(t) = x(t), α ̸= 0, is given}

by y(t) = (5 + 3e−2t_)u(t).

(a) Identify the natural and forced response. (b) Identify the values of α and y(0).

4.25 (System Response) It is known that the response of the system y′_{(t) + y(t) = x(t) is given by} y(t) = (5e−t+ 3e−2t)u(t).

(a) Identify the zero-input and zero-state response.

(b) What is the zero-input response of the system y′_{(t) + y(t) = x(t) if y(0) = 10?} (c) What is the response of the relaxed system y′_{(t) + y(t) = x(t − 2)?}

(d) What is the response of the relaxed system y′_{(t) + y(t) = x}′_{(t) + 2x(t)?}

4.26 (System Response) It is known that the response of the system y′_{(t) + αy(t) = x(t) is given by} y(t) = (5 + 2t)e−3tu(t).

(a) Identify the zero-input and zero-state response.

(b) What is the zero-input response of the system y′_{(t) + αy(t) = x(t) if y(0) = 10?} (c) What is the response of the relaxed system y′_{(t) + αy(t) = x(t − 2)?}

(d) What is the response of the relaxed system y′_{(t) + αy(t) = 2x(t) + x}′_(t)?

Chapter 4 Problems 93 4.27 (Impulse Response) Consider the relaxed system y′_{(t) +}1

τy(t) = x(t). (a) What is the impulse response of this system?

(b) What is the response of this system to the rectangular pulse x(t) = 1

α[u(t) − u(t − α)]? Show that as α → 0, we obtain the system impulse response h(t).

αe−t/αu(t)? Show that as α→ 0, we obtain the system impulse response h(t).

4.28 (System Response) Find the response of the following systems for t ≥ 0. (a) y′_{(t) + 2y(t) = 2e}−(t−1)_u(t_{− 1)} _{y(0) = 5}

(b) y′_{(t) + 2y(t) = e}−2t_{u(t) + 2e}−(t−1)_u(t_{− 1)} _{y(0) = 5} (c) y′_{(t) + 2y(t) = te}−t_{+ 2e}−(t−1)_u(t_{− 1)} _{y(0) = 5} (d) y′_{(t) + 2y(t) = cos(2t) + 2e}−(t−1)_u(t_{− 1)} _{y(0) = 5}

4.29 (Impulse Response) Find the step response and impulse response of each circuit in Figure P4.29.

y(t) x(t) x(t) x(t) y(t) y(t) y(t) y(t) x(t) x(t) x(t) y(t)

Circuit 4 Circuit 5 Circuit 6

+ + − − R L + R − C R

Circuit 1 Circuit 2 Circuit 3

R + − − R R C + + − − + + − − R C R + − R + L + − + −

Figure P4.29 Circuits for Problem 4.29

4.30 (Impulse Response) The input-output relation for an LTI system is shown in Figure P4.30. What is the impulse response h(t) of this system?

Input t 4 x(t) 1 3 2 4 1 3 −4 5 t y(t) Output

Figure P4.30 Figure for Problem 4.30

4.31 (System Response) Consider two relaxed RC circuits with τ1 = 0.5 s and τ2 = 5 s. The input to both is the rectangular pulse x(t) = 5 rect(t − 0.5) V. The output is the capacitor voltage.

(a) Find and sketch the outputs y1(t) and y2(t) of the two circuits.

(b) At what time t > 0 does the output of both systems attain the same value?

4.32 (Classification and Stability) Argue for or against the following statements, assuming relaxed systems and constant element values. You may validate your arguments using simple circuits.

(a) A system with only resistors is always instantaneous and stable. (b) A system with only inductors and/or capacitors is always stable.

94 Chapter 4 Analog Systems 4.33 (Differential Equations from Impulse Response) Though there is an easy way of obtaining a system differential equation from its impulse response using transform methods, we can also obtain such a representation by working in the time domain itself. Let a system be described by h(t) = e−t_{u(t). If} we compute h′_{(t) = δ(t) −e}−t_{u(t), we find that h}′_{(t) + h(t) = δ(t), and the system differential equation} follows as y′_{(t) + y(t) = x(t). Using this idea, determine the system differential equation corresponding} to each impulse response h(t).

(a) h(t) = e−αt_u(t) _{(b) h(t) = e}−t_u(t)_{− e}−2t_u(t)

4.34 (Inverse Systems) If the input to a system is x0(t) and its response is y0(t), the inverse system is defined as a system that recovers x0(t) when its input is y0(t). Inverse systems are often used to undo the eﬀects of measurement systems such as a transducers. The system equation of the inverse of many LTI systems can be found simply by switching the input and output. For example, if the system equation is y(t) = x(t −3), the inverse system is x(t) = y(t−3) (or y(t) = x(t+3), by time invariance). Find the inverse of each system and determine whether the inverse system is stable.

(a) y′_{(t) + 2y(t) = x(t)} _{(b) y}′′_{(t) + 2y}′_{(t) + y(t) = x}′_{(t) + 2x(t)}

4.35 (Inverse Systems) A requirement for a system to have an inverse is that unique inputs must produce unique outputs. Thus, the system y(t) = |x(t)| does not have an inverse because of the sign ambiguity. Determine which of the following systems are invertible and, for those that are, find the inverse system.

(a) y(t) = x2_(t) _{(b) y(t) = e}x(t) _{(c) y(t) = cos[x(t)]} (d) y(t) = ejx(t) _{(e) y(t) = x(t − 2) (f) y}′_{(t) + y(t) = x(t)}

COMPUTATION AND DESIGN

ctsimgui

A GUI for Numerical Simulation of System Response

The graphical user interface ctsimgui allows you to simulate and visualize the response of analog systems, using various numerical integration methods.

You can select the input signal, the system parameters, and the simulation method. You can display the input and simulated response for various choices of the time step. To explore this routine, type ctsimgui at the Matlab prompt.

4.36 (System Response in Symbolic Form) The ADSP routine sysresp1 yields a symbolic result for the system response (see Chapter 21 for examples of its usage). Consider the system y′_{(t) + 2y(t) =} 2x(t). Use sysresp1 to obtain its

(a) Step response. (b) Impulse response.

(d) Complete response to x(t) = 4e−3t_{u(t) with y(0) = 5.}

4.37 (System Response) Use the ADSP routine sysresp1 to find the step response and impulse response of the following filters and plot each result over 0 ≤ t ≤ 4. Compare the features of the step response of each filter. Compare the features of the impulse response of each filter.

(a) y′_{(t) + y(t) = x(t)} _{(a first-order lowpass filter)}

Chapter 4 Problems 95 (c) y′′_{(t) + y}′_{(t) + y(t) = x}′_(t) _{(a bandpass filter)}

(d) y′′′_{(t) + 2y}′′_{(t) + 2y}′_{(t) + y(t) = x(t)} _{(a third-order Butterworth lowpass filter)}

4.38 (Rise Time and Settling Time) For systems whose step response rises to a nonzero final value, the rise time is commonly defined as the time it takes to rise from 10% to 90% of the final value. The settling time is another measure for such signals. The 5% settling time, for example, is defined as the time it takes for a signal to settle to within 5% of its final value. For each system, use the ADSP routine sysresp1 to find the impulse response and step response and plot the results over 0 ≤ t ≤ 4. For those systems whose step response rises toward a nonzero final value, use the ADSP routine trbw to numerically estimate the rise time and the 5% settling time.

(a) y′_{(t) + y(t) = x(t)} _{(a first-order lowpass filter)}

(b) y′′_{(t) +}√_2y′_{(t) + y(t) = x(t)} _{(a second-order Butterworth lowpass filter)} (c) y′′_{(t) + y}′_{(t) + y(t) = x}′_(t) _{(a bandpass filter)}

(d) y′′′_{(t) + 2y}′′_{(t) + 2y}′_{(t) + y(t) = x(t)} _{(a third-order Butterworth lowpass filter)} 4.39 (System Response) Consider the system y′′_{(t) + 4y(t) + Cy(t) = x(t).}

(a) Use sysresp1 to obtain its step response and impulse response for C = 3, 4, 5 and plot each response over an appropriate time interval.

(b) How does the step response diﬀer for each value of C? For what value of C would you expect the smallest rise time? For what value of C would you expect the smallest 3% settling time? (c) Confirm your predictions in the previous part by numerically estimating the rise time and settling

time, using the ADSP routine trbw.

4.40 (Steady-State Response in Symbolic Form) The ADSP routine ssresp yields a symbolic ex- pression for the steady-state response to sinusoidal inputs (see Chapter 21 for examples of its usage). Find the steady-state response to the input x(t) = 2 cos(3t −π

3) for each of the following systems and plot the results over 0 ≤ t ≤ 3, using a time step of 0.01 s.

(a) y′_{(t) + αy(t) = 2x(t), for α = 1, 2}

(b) y′′_{(t) + 4y(t) + Cy(t) = x(t), for C = 3, 4, 5}

4.41 (Numerical Simulation of Analog Systems) The ADSP routine ctsim returns estimates of the system response using numerical integration such as Simpson’s rule and Runge-Kutta methods. Consider the diﬀerential equation y′_{(t) + αy(t) = αx(t). In the following, use the second-order Runge-} Kutta method throughout.

(a) Let x(t) = rect(t − 0.5) and α = 1. Evaluate its response y(t) analytically. Use ctsim to evaluate its response y1(t) over 0 ≤ t ≤ 3, using a time step of 0.1 s. Plot both results on the same graph and compare.

(b) Let x(t) = sin(t), 0 ≤ t ≤ π. Use ctsim to evaluate its response y1(t) over 0 ≤ t ≤ 6, using α = 1, 3, 10 and a time step of 0.02 s. Plot each response along with the input x(t) on the same graph. Does the response begin to resemble the input as α is increased? Should it? Explain. (c) Let x(t) = sin(t), 0 ≤ t ≤ π. Use ctsim to evaluate its response y1(t) over 0 ≤ t ≤ 6, using

α = 100 and a time step of 0.02 s. Plot the response along with the input x(t) on the same graph. Now change the time step to 0.03 s. What is the response? To explain what is happening, find and plot the response for time steps of 0.0201 s, 0.0202 s, and 0.0203 s. Describe what happens to the computed response and why.

Chapter 5 DISCRETE-TIME SYSTEMS

5.0 Scope and Objectives

The processing of discrete-time signals is accomplished by discrete-time (DT) systems or digital filters. Their description relies heavily on how they respond to arbitrary or specific signals. Many discrete-time systems may be described by diﬀerence equations relating the input and output. The class of linear, time- invariant systems may also be described by their impulse response, the response to an impulse input. This chapter deals with discrete-time systems and their classification, time-domain representation, and analysis based on the solution of diﬀerence equations. It also introduces the all-important concept of the impulse response, which forms a key ingredient in both system description and system analysis.

In document Analog and Digital Signal Processing (Page 100-110)