Mid-term Exam ECE 3EJ4 17 Oct 2013

Answer all 4 questions. Circle the letters e.g. (A) corresponding to your answers. Time – 120 minutes. 1. The following is a Bode plot of an amplifier circuit with 3 poles and 3 zeros.

Frequency (Hz)

60 65 70 75 80 85 90 95 100 Magnitude (dB) Bode Diagram

10-1 100 101 102 103 104 105 106 107

Exact Approximation

a. What is the mid-band gain of this system (in dB)?

(a) 60 (b) 102 (c) 103 (d) 104 (d) 1 (e) 80 (f) 105 b. Find all the pole and zero frequencies of this system.

(a) pole frequencies (rad/s): 10, 105, 106; zero frequencies (rad/s): 102_{, 10}2_{, 10}6

(b) pole frequencies (rad/s): 102, 105, 106; zero frequencies (rad/s): 10, , 106, 106

(c) pole frequencies (rad/s): 10, 10, 106; zero frequencies (rad/s): 102_{, 10}2_{, 10}5

(d) pole frequencies (rad/s): 102, 102, 105; zero frequencies (rad/s): 10, 10, 106

(e) pole frequencies (rad/s): 102, 105, 105; zero frequencies (rad/s): 10, 106_{, 10}6

c. Write the transfer function of this system.

(a)

 















2 6

3

2 5

1 10 1 10

10

1 10 1 10

s s

H s

s s

 



  (b)

 















2 6

4

2 5

1 10 1 10

10

1 10 1 10

s s H s s s     

(c)

 











 



2 6

3

2

2 5

1 10 1 10

10

1 10 1 10

s s

H s

s s

 



  (d)

 















6 3

2 5

1 10 1 10

10

1 10 1 10

s s H s s s     

(e)

 















6 4

2 5

1 10 1 10

10

1 10 1 10

s s

H s

s s

 



  (f)

 











 



2 6 4 2 2 5

1 10 1 10

10

1 10 1 10

s s H s s s     

(g)

 











 



2 6

2

2

2 5

1 10 1 10

10

1 10 1 10

s s

H s

s s

 



  (h)

 











 



6 5 2 2 5

1 10 1 10

10

1 10 1 10

s s H s s s     

(i)

 















2 6

2

2 5

1 10 1 10

10

1 10 1 10

s s

H s

s s

 



  (j)

 











 



2 6 5 2 2 5

1 10 1 10

10

1 10 1 10

d. What is the approximate phase shift at  = 10-2 rad/s from the asymptotic Bode plot?

(a) 0˚ (b) 45˚ (c) 90˚ (d) 135˚ (e) 180˚

(f) -45˚ (g) -90˚ (h) -135˚ (i) -180˚

e. What is the approximate phase shift at  = 5103 rad/s from the asymptotic Bode plot?

(a) 0˚ (b) 45˚ (c) 90˚ (d) 135˚ (e) 180˚

(f) -45˚ (g) -90˚ (h) -135˚ (i) -180˚

f. What is the approximate phase shift at  = 108 rad/s from the asymptotic Bode plot?

(a) 0˚ (b) 45˚ (c) 90˚ (d) 135˚ (e) 180˚

(f) -45˚ (g) -90˚ (h) -135˚ (i) -180˚

(Hint: Draw the Bode diagram for the phase)

10-2 10-1 100 101 102 103 104 105 106 107 108 -180

-135 -90 -45 0 45 90 135 180

Frequency (Hz)

pha

s

e

(

d

e

g

)

g. What is the bandwidth in rad/s?

2. The circuit of a common-gate amplifier (left) and it’s simple small-signal model in open-loop (right) are shown below.

a. What feedback type is this circuit?

(a) V/V (b) I/I (c) V/I (d) I/V

b. What is β?

 

1 2

D

D R a

R R R

 

1 2

RD b

R R

 

2

1 2

R

c

R R

 

1

1 2

R

d

R R

 

1

1 2

D R e

R R R

 

1 2

1 2

R R

f

R R

 

2

1 2

D R g

R  R R

 





1

1 2

//

D R g

R R R

c. What is the open-loop gain when taking into account the effect of R1 and R2?

 

a g R_{m D}

 

b gm(R1R2)

 

c //(gm



RD R1R2)



d. What is the closed-loop gain? (a) 2 1 2 1 m D m D g R R g R R R  

(b)









1 2 2 1 2 1 2 //( )

1 //( )

m D

m D

g R R R

R

g R R R

R R



 



(c)









1 2 2 1 2 1 2 ( // )

1 ( // )

m D

m D

g R R R

R

g R R R

R R



 



(d) 1 2

2

1 2

1 2

( // )

1 ( // )

m

m

g R R

R

g R R

R R   (e) 2 1 1 m D m D g R R g R R 

(f)









1 2 2 1 2 1 //( )

1 //( )

m D

m D

g R R R

R

g R R R

R



 

(g)









1 2 2 1 2 1 ( // )

1 ( // )

m D

m D

g R R R

R

g R R R

R



  (h)

1 2

2

1 2

1

( // )

1 ( // )

m

m

g R R

R

g R R

R



e. What is the open-loop input resistance?

(a) g_m (b)  (c) 0 (d) 1

m

g

(e) R₁//R₂ (f) R_D (g) 1 m

r

g  

f. What is the closed-loop input resistance?

(a) 2





1 2 1 1 ( // ) D m R

R R R

g  R  (b)





2 1 2 1 2 1 ( // ) D m R

R R R

g R R 

(c) 2





1 2 1 1 //( ) D m R

R R R

g  R  (d)





2 1 2 1 2 1 //( ) D m R

R R R

g  R R 

(e) 2





1 2

1 1

( // ) 1

D m D

D

m D m D

R g R R

R R R

g R  R g R R 

  (f)





2

1 2

1 2

( // )

1 ( )(1 )

D m D

D

m D m D

R g R R

R R R

g R  R R g R 

  

(g) 2





1 2

1 1

//( )

1

D m D

D

m D m D

R g R R

R R R

g R R R g R 

  (h)





2

1 2

1 2

//( )

1 ( )(1 )

D m D

D

m D m D

R g R R

R R R

g R  R R g R 

g. What is the open-loop output resistance?

(a) R₂ (RD// )R₁ (b) ro ( //R1 R2) (c) 0 (d)

1

m

g

(e) R₁//R₂ (f) R_D (g)  (g) R_D//( // )R₁ R₂

(h) R_D//( + )R R_{1 2} (i) r_o( //R₁ R_D)

h. What is the closed-loop output resistance?

(a)





1 2

2

1 2

1

( // )

1 ( // )

D

m D

R R R

g R

R R R

R



 

(b)





1 2

2

1 2

1

//( )

1 //( )

D

m D

R R R

g R

R R R

R



 

(c)





1 2

2

1 2

1 2

//( )

1 //( )

D

m

D

R R R

g R

R R R

R R



 



(d)





1 2

2

1 2

1 2

( // )

1 ( // )

D

m

D

R R R

g R

R R R

R R



 



(e)





1 2

2

1 2

1 1

( // )

1 ( // )

D

m D

D

m D

R R R

g R R

R R R

R g R R



 



(f)





1 2

2

1 2

1 1

//( )

1 //( )

D

m D

D

m D

R R R

g R R

R R R

R g R R



 

3. Regarding the circuit shown below, assume the amplifier is ideal.

a. If this circuit is a second-order low pass filter, which

passive components should Z1, Z2, Z3 and Z4 be respectively?

(a) resistor, resistor, capacitor, capacitor (b) capacitor, resistor, capacitor, resistor (c) resistor, capacitor, capacitor, resistor (d) capacitor, resistor, resistor, capacitor (e) capacitor, resistor, capacitor, capacitor (f) resistor, capacitor, capacitor, capacitor (g) capacitor, resistor, resistor, resistor (h) capacitor, capacitor, resistor, resistor (i) resistor, capacitor, resistor, capacitor

b. If Rf = 0 and R1 = ∞, what is the low frequency gain of the circuit?

(a) 0 (b) -0.5 (c) -1 (d) -1.5

(e) -2 (f) -10 (g) 0.5 (h) 1

(i) 1.5 (j) 2 (k) 10

c. Regarding the second-order low pass filter, assuming |Z1|=|Z3|, |Z2|=|Z4|, and using the resistor of

value 10k and the capacitor of value 1nF, what is the pole frequency (fo) in Hz?

(a) 628 kHz (b) 314 kHz (c) 100 kHz (d) 64 kHz

d. Calculate the quality factor Q based on previous questions

(a) 0 (b) 0.5 (c) 0.707 (d) 1

(e) 2 (f) 5 (g) 10 (h) 20

(i) 50 (j) 100

e. Assuming conditions of Question 3(c), what is the phase shift at ωo?

(a) 0° (b) 90° (c) 110.10° (d) 150.45°

(e) 168.58° (f)178.8° (g) -90° (h)-110.10°

(i) -150.45° (j) -168.58° (k) -178.8°

f. Assuming conditions of Question 3(c), what is the phase shift at 10ωo?

(a) 0° (b) 90° (c) 110.1° (d) 150.5°

(e) 168.6° (f)178.8° (g) -90° (h)-110.1°

(i) -150.5° (j) -168.6° (k) -178.8°

g. Assuming conditions of Question 3, what is the phase shift at 100ωo?

(a) 0° (b) 90° (c) 110.1° (d) 150.5°

(e) 168.6° (f)178.8° (g) -90° (h)-110.1°

4. A Wien-Bridge oscillator circuit with frequency of oscillation fo=1kHz and a power supply voltage ±15V.

The 2 diodes are identical and can modeled with: VD,on=0.7V and RD=1kΩ.

a. What is the value of C?

(a) 160 nF (b) 16n F (c) 0.16 nF (d) 1.6 nF (e) 100 µF (f) 10 µF (g) 1 µF

b. If both R and C are reduced by half, what would be the new fo?

(a) 1 kHz (b) 2 kHz (c) 4 kHz (d) 0.5 kHz

(e) 0.25 kHz (f) 25 kHz (g) 6.25 kHz (h) 12.5 kHz

c. Write an expression for vo (in volts):

(a) 4.8×sin(2πfot) (b) 5.6×sin(2πfot) (c) 15.6×sin(2πfot) (d) 6.5×sin(2πfot) (e) 6.5×cos(2πfot) (f) 30×sin(2πfot) (g) 13×sin(2πfot) (h) 11.2×sin(2πfot) (i) 5×sin(2πfot) (j) 10×sin(2πfot)

d. Write an expression for vo (in volts), if the diodes are replaced by others which can be modeled this

time with only VD,on=0.7V (that is, no RD in the diode model):

4. A PLL has fi=100MHz, fosc=95MHz, and osc=45o.

e. Write an expression for Vd:

( ) [sin(390 45 ) sin(10o 45 )]o D

a K M t   M t 

( ) [sin(390 45 ) sin(10 45 )]

2

o o

i osc

V V

b M t   M t 

( ) [sin(390 45 ) sin(10o 45 )]o D i osc

c K V V M t   M t 

( ) [sin(390 45 ) sin(10 45 )]

2

o o

D i osc

K V V

d M t   M t 

( ) [sin(195 45 ) sin(5o 45 )]o D i osc

e K V V M t   M t 

( ) [sin(390 45 ) sin(10o 45 )]o D i osc

f K V V M t   M t 

( ) [sin(390 45 ) sin(10 45 )]

2

o o

D i osc

K V V

g M t   M t 

f. If the PLL is tuned so that it locks at an FM radio station with frequency 90MHz, what is fi and fosc

in this case?

(a) 180π MHz, 90 MHz (b) 90 MHz, 180π MHz (c) 100 MHz, 90 MHz (d) 90 MHz, 10 MHz (e) 90 MHz, 90 MHz (f) 100 MHz, 0 (g) 90 MHz, 0 (h) 0, 90 MHz

g. What is the value of Ve at lock conditions if F(s) = 1 osc=90o?

(a) 0.5KDViVoscF(s) (b) KDViVosc (c) 0.5KDViVosc (d) 0.5Vosc/Ko (e) KDVi

Useful Information – remove this page for easy reference.

LP – Low-pass HP – High pass BP – Band pass BS – Band stop or Notch AP – All pass

Series Impedance _series 



_k

k

Z Z Parallel Impedance _{1 2} 1 2

1 2 // 1 + 1 paral parallel

lel n n

Z Z

Z Z Z

e.g. Z Z Z Z    



Common-source single-stage FET amplifier with load resistance RL is shown

Voltage Gain A_v  g_m



R_D//R_L//r_o



. ro is the ss-resistance between D & S

Kirchoff’s Current Law (KCL)



n =0



_in =



_o

n

ut

i i

i or

Kirchoff’s Voltage Law (KVL)



 =0



_{sources ,m} =



_n _n

m n n n V i or Z V

Capacitor  1  1  C

Z

sC j C InductorZL sL j L

Complex : _{   } 2_ 2_{;     } 1 _{; }  _



_{  }



 

  

j

y

r x jy r x y r tan r r e r cos j sin

x

 

 

 

1 ₁₀ 2 _{0 57 ;} o 1 ₁₀ 1 _{5 7 ;} o 1 _{1 45}o

tan   . tan   . tan 

 

 

1 _{10 84 3 ;} o 1 _{100 89 42}o

tan  . tan  .

Voltage Amplifier Vin-Vout Series-Shunt Feedback R_if R_i



1 A



; R_of R_o



1 A



Current Amplifier Iin-Iout Shunt-Series Feedback R_if R_i



1 A



; R_of R_o



1 A



Transconductance Amplifier Vin-Iout Series- Series Feedback R_if R_i



1 A



; R_of R_o



1 A



Transresistance Amplifier Iin-Vout Shunt –Shunt Feedback



1



;



1



if i of o

R R  A R R  A

Simple rule to determine loading effect of feedback network on basic amplifier. Shunt connection  Short the port

Series connection  Sever or open the port

Example: For the voltage amplifier shown, this means that

port 1 is severed looking from output (R2+ R1 connected to output at 2), and

port 2 is shorted looking from input (R1 // R2 connected to input 1 or minus

terminal of Op-amp)

Transconductance gm with source resistor RS degeneration

1   m m,eff m S g g g R Small-signal models FET and BJT respectively C C vd D G 2 1 1    f A A A

 

2 0 2 1 2 0 1 1 2 0 2 0 2 0 0 0 ; 0 LP

a s a s a