EE 179 April 21, 2014 Digital and Analog Communication Systems Handout #16 Homework #2 Solutions

EE 179 April 21, 2014

Digital and Analog Communication Systems Handout #16

Homework #2 Solutions

1. Operations on signals (Lathi & Ding 2.3-3). For the signal g(t) shown below, sketch: a. g(t−4); b. g(t/1.5); c. g(2t−4); d. g(2−t).

Hint: Recall that replacing t with t−T delays the signal by T. Thus g(2t−4) is g(2t) with t

replaced by t−2. Similarly, g(2−t) is g(−t) with t replaced byt−2. Solution _{(20 points)}

2. Orthogonal signals (Lathi & Ding 2.5-5). The energies of two energy signals x(t) and y(t) are Ex and Ey, respectively.

a. If x(t) and y(t) are orthogonal, then show that the energy of the signalx(t) +y(t) is identical to the energy of the signal x(t)−y(t) and is given byEx+Ey.

b. Ifx(t) andy(t) are orthogonal, find the energies of the signalsc1x(t)+c2y(t) andc1x(t)−c2y(t). c. We define Exy, the correlation of the two energy signals x(t) and y(t), as

Exy = Z ∞

−∞

x(t)y∗(t)dt .

If z(t) =x(t)±y(t), then show that

Ez =Ex+Ey ±(Exy +Eyx). Solution _{(20 points)}

a. If x(t) and y(t) are orthogonal, then by definition Z ∞ −∞ x(t)y∗(t)dt= 0 and Z ∞ −∞ x∗(t)y(t)dt= Z ∞ −∞ x(t)y∗(t)dt∗ = 0..

Therefore Z ∞ −∞ |x(t)±y(t)|2dt= Z ∞ −∞ (x(t)±y(t))(x(t)±y(t))∗dt = Z ∞ −∞ |x(t)|2dt± Z ∞ −∞ x(t)y∗(t)dt± Z ∞ −∞ x∗(t)y(t)dt+ Z ∞ −∞ |y(t)|2dt = Z ∞ −∞ |x(t)|2dt+ Z ∞ −∞ |y(t)|2dt =Ex+Ey. b. Since c1x(t) and c2y(t) are orthogonal,

Ec1x±c2y =Ec1x+Ec2y =|c1|

2_E

x+|c2|2Ey. c. By part (a), if z(t) =x(t)±y(t) then

Ez =Ex+Ey ±(Exy +Eyx).

Note that Exy +Eyx = Exy +Exy∗ = 2 ReExy. If x(t) and y(t) are orthogonal, then Exy = 0 and therefore Ez = Ex+Ey. Consider z(t) = x(t) +y(t). If ReExy < 0 then Ez < Ex +Ey since x(t) andy(t) partially cancel each other. If ReExy >0 thenEz < Ex+Ey since x(t) and

y(t) reinforce.

3. Use Fourier transform properties to derive Fourier transforms (Lathi & Ding 3.3-2). The Fourier transfer of the triangular pulse g(t) in Fig. P3.3-2a is given as

G(f) = 1 (2πf)2 e

j2πf _{−j2πf e}j2πf ₋₁

.

Use this information, and time-shifting and time-scaling properties, to find the Fourier transforms of the signals shown is Fig. P3.3-2b, c, d, e, and f.

Solution _{(20 points)}

a. By the problem statement,

G(f) = 1 (2πf)2 e

j2πf _{−j2πf e}j2πf ₋₁

b. Since g1(t) =g(−t),G1(f) =G(−f): G1(f) = 1 (2πf)2 e −j2πf _{+j2πf e}−j2πf₋₁ .

c. g2(t) =g(t−1) +g1(t−1). Then by the shift theorem,

G2(f) = e −j2πf (2πf)2 e j2πf _{−j2πf e}j2πf₋₁ + e −j2πf (2πf)2 e −j2πf _{+j2πf e}−j2πf₋₁ = 1 (2πf)2 1−j2πf −2e −j2πf +e−j4πf +j2πf e−j4pif . d. g3(t) =g(t−1) +g1(t+ 1). Then G3(f) = e −j2πf (2πf)2 e j2πf _{−j2πf e}j2πf₋₁ + e j2πf (2πf)2 e −j2πf _{+j2πf e}−j2πf₋₁ = 1 (2πf)2 2−e −j2πf _−ej2πf = 1 (2πf)2 (2−2 cos 2πf). Another formula for G3(f) is sinc2f.

e. g4(t) =g(t− 1₂) +g1(t+1₂). Then G4(f) = e −jπf (2πf)2 e j2πf _{−j2πf e}j2πf₋₁ + e jπf (2πf)2 e −j2πf _{+j2πf e}−j2πf₋₁ = 1 (2πf)2 e jπf _{−j2πf e}jπf _−e−jπf_+e−jπf _{+j2πf e}−jπf_−ejπf = 1 (2πf)2 −j2πf e jπf _{+j2πf e}−jπf = −j 2πf e jπf _−e−jπf = sinπf πf f. g5(t) = 3₂g(1₂(t−2)). Then G5(f) = 3₂e−j4πf2G(2f) = 3e −j4πf (4πf)2 e j4πf_−j4πej4πf₋₁ = 3 (4πf)2 1−j4πf −e −j4πf .

4. Modulation and demodulation.

a. Let m(t) be a message signal, fc a constant (carrier frequency), and define

x(t) =m(t) cos(2πfct+θ0).

LetM(f) be the Fourier transform ofm(t). FindX(f), the Fourier transform ofx(t), in terms of M(f).

b. Find the signal (in the time domain), whose Fourier transform is pictured below.

✦✻ ✦✹ ✦✷

✵ ✶

c. A similar relationship can be found for x(t) = m(t) sin(2πfct). Find it, and use it to find the Fourier transform of x(t) = ( sin(2πt) |t|<1/2 0 |t|>1/2

without performing any integration. Does the Fourier transform have the properties you would expect (even/odd/neither, real/imaginary/complex)?

d. Show that m(t) can be recovered from x(t) = m(t) cos(2πfct) by multiplying by 2 cos(2πfct) and passing the product through a low-pass filter of bandwidthB Hz, whereB is the bandwidth of m(t). Assume that B ≪fc.

Solution _{(20 points)}

a. We appeal directly to the definition of the Fourier transform:

X(f) = Z ∞ −∞ x(t)e−j2πf t_dt = Z ∞ −∞ m(t) cos(2πfct+θ0)e−j2πf tdt = Z ∞ −∞ m(t)1₂(ej2πfct+θ0 +e−j2πfct−θ0 )e−j2πf tdt = 1 2 Z ∞ −∞ m(t)e−j2π(f−fc)t+θ0 dt+1 2 Z ∞ −∞ m(t)e−j2π(f+fc)t−θ0 dt = 1 2M(f−fc)e jθ0 +1 2M(f +fc)e −jθ0

b. With an eye toward using part (a) with θ0 = 0, the Fourier transform in this part can be written as 1 2 ·2∆ f −4 4 + 1₂ ·2∆ f + 4 4 .

where ∆(f) is the triangle function with height 1 and width 2. So it looks like we want to find a function m(t) whose Fourier transform is 2∆(f /4), for if we then multiply it by cos(2π·4·t) the modulation property gives us what we want for the Fourier transform:

F {m(t) cos 8πt}= ∆ f−4 4 + ∆ f −4 4 .

Since F {∆(t/τ)}=τsinc2(τ f), we obtain by duality F τsinc2(τ t) = ∆(f /τ), If we thus set m(t) = 4 sinc2(2t) we have M(f) = 2∆(f /4)

as desired. So, the signal x(t) whose Fourier transform X(f) in the picture is

c. The computation is very similar to what was done in part (a): X(f) = Z ∞ −∞ x(t)e−j2πf tdt = Z ∞ −∞ m(t) sin(2πfct)e−j2πf tdt = Z ∞ −∞ m(t) 1 2j(e j2πfct −e−j2πfct )e−j2πf tdt = 1 2j Z ∞ −∞ m(t)e−j2π(f−fc)t dt− 1 2j Z ∞ −∞ m(t)e−j2π(f+fc)t dt = 1 2jM(f−fc)− 1 2jM(f +fc)

(We can also findX(f) using the convolution theorem: X(f) =M(f)∗ 1

2j δ(f−fc)−δ(f+fc) , where 1 2j δ(f −fc)−δ(f +fc)

is the Fourier transform of sin(2πfct).) To find the Fourier transform of x(t) = ( sin(2πt) |t| ≤1/2 0 |t| ≥1/2 we note that x(t) = Π(t) sin 2πt

and we can apply the sine-modulation formula with fc = 1. Since F {Π(f)}= sinc(f),

X(f) = 1

2j sinc(f −1)−sinc(f+ 1)

.

d. The modulated signal is x(t) =m(t) cos(2πfct). Multiplying by 2 cos(2πfct) yields 2m(t) cos2(2πfct) =m(t) 1 + cos(4πfct)

=m(t) +m(t) cos(4πfct).

Observe that the resulting signal contains the original signal m(t) and a modulated copy of the signal moved to a frequency center of 2fc. If the bandwidth of the original signal isB < fc, then the modulated copy will not extend further than fc from its center frequency and a low-pass filter from −fc to +fc will pass only m(t) and filter out the modulated copy.

5. Essential bandwidth (Lathi & Ding 3.7-4). For the signal

g(t) = 2a

t2_+a2

determine in hertz the essential bandwidthB ofg(t) such that the energy contained in the spectral components of g(t) of frequencies below B Hz is 99% of the signal energy Eg. Hint: determine

G(f) by applying the duality property [Eq. (3.26)] to pair 3 of Table 3.1. Solution _{(10 points)}

Applying the duality property [Eq. (3.26)] to pair 3 of Table 3.1, we find F 2a a2_{+ (2πt)}2 =e−a|f|.

Applying the time-scaling property with the constant _2π1 , we obtain F 2a a2_+t2 = 2πe−a|2πf|.

The signal energy is easily calculated in the frequency domain:

Eg = Z ∞ −∞ |G2(f)|df = Z ∞ −∞ |2πe−a|2πf||2df = 2·4π2 Z ∞ 0 e−4aπfdf = 8π2 1 4aπ = 2π a .

The energy of the signal contained in the high-pass band beyond B Hz is

EHF = 2·4π2 Z ∞ B e−4aπf df = 2π a e −4aπB_.

Setting EHF= 0.01Eg and solving for B, we obtain 2π a e −4aπB _{= 0.01·}2π a ⇒ e −4aπB _{= 0.01 ⇒ B =} ln(0.01) −4aπ = ln(100) 4aπ .

6. Autocorrelation and PSD. Show that the autocorrelation function of g(t) = Acos(2πf0t+θ0) is

Rg(τ) = 1₂A2cos(2πf0τ) and that the corresponding PSD isSg(f) = 1₄A δ(f−f0) +δ(f+f0)

. Solution _{(10 points)}

Using the definition of the autocorrelation function for power signals as given in Eq. (3.83a),

Rg(τ) = lim T→∞ 1 T Z T /2 −T /2 g(t)g(t−τ)dτ = lim T→∞ A2 T Z T /2 −T /2 cos(2πf0t+θ0) cos(2πf0t−2πf0τ +θ0)dt = lim T→∞ A2 2T Z T /2 −T /2 cos(2πf0τ)dt+ A2 2T Z T /2 −T /2 cos(4πf0t−2πf0τ + 2θ0)dt !

(using trigonometric identity cosacosb= 1₂(cos(a−b) + cos(a+b))) = lim T→∞ A2 2T T cos(2πf0τ) + A2 2T sin(2πf0(T +τ)−2θ0) + sin(2πf0(T −τ) + 2θ0) 4πf0 = A 2 2 cos(2πf0τ) + limT→∞ A2 2T f(t) 4πf0 = A 2 2 cos(2πf0τ).

The limit of the second term in the fifth line is 0 because the numeratorf(t) is bounded in absolute value by 2. The power spectral density of the signal is the Fourier transform of Rg(τ):

Sg(f) = A 2

2 δ(f +f0) +δ(f −f0)

7. Output SNR of linear system(Lathi & Ding 3.8-5). Consider a linear system with impulse response

e−2t_{u(t). The linear system input is}

g(t) = w(t)−cos6πt+π 3

,

in which w(t) is a noise signal with power spectral density Sw(f) = Π(f /4). a. Find the total output power of the linear system.

b. Find the output power of the signal component due to the sinusoidal input. c. Find the output power of the noise component.

d. Determine the output signal-to-noise ratio (SNR) in decibels. Solution _{(20 points)}

a. The power spectral density of the sinusoidal signal is Sx(f) = 1₄(δ(f+ 3) +δ(f−3)), hence the input PSD is

Sg(f) = Sw(f) +Sx(f) = Π(f /4) + 1₄(δ(f+ 3) +δ(f −3)).

The transfer function of the system is the Fourier transform of the impulse response: F e−2tu(t) = 1 2 +j2πf ⇒ |H(f)| 2 ₌ 1 4 + 4π2_f2 . The output PSD is Sy(f) =|H(f)|2Sg(f) = 1 4 + 4π2_f2 Π(f /4) + 1 4(δ(f + 3) +δ(f−3)) .

and so the output power is

Py = Z ∞ −∞ Sy(f)df = Z 2 −2 1 4 + 4π2_f2df + 1/4 4 + 4π2₍₋₃₎2 + 1/4 4 + 4π2₍₊₃₎2 = tan −1_(2π) 2π + 1 8 + 72π ≈0.22915.

b. The PSD of the output due to the sinusoidal signal is

Syx(f) = |H(f)|2Sx(f) = 1 4 + 4π2_f2 1 4(δ(f+ 3) +δ(f −3)) ,

and the total power due to the signal is

Pyx= Z ∞

−∞

Syx(f)df = 1

8 + 72π ≈0.00427.

c. The power due to the noise is

Pyw = Z ∞ −∞ Syw(f)df = Z 2 −2 1 4 + 4π2_f2 df = tan−1_(2π) 2π ≈0.22488.

d. The signal-to-noise ratio in decibels is SNR = 10 log₁₀ Pyx

Pyw

≈10 log₁₀0.00427

0.22488 ≈ −17.2 dB

Note that the negative SNR means that the power of the signal is less than the power of the noise.