EEE307 Test Your Understanding

大学学习

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Test Your Understanding

Author: BYJR_K©

P.S. The Online LaTeX Editor can be helpful.

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Question 1

30-37

Electrical Noise

What is electrical noise? And what are the main external noise and internal noise in communication systems? (exercise questions)

Electrical noise

Electrical noise may be defined as any undesired voltages or currents that ultimately end up appearing in the receiver output.

External noise - in transmitting medium

1. Human-made noise - engine ignition systems, communicator in electric motors

2. Atmospheric noise - lighting in the earth atmosphere

3. Space noise - solar noise from the sun, reaches peaks about every 11 years

Internal noise - in receiver

1. Thermal noise(热杂波) - resistor

2. Transistor noise - transistor (amplifier)

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Question 2

42

Noise Ratio/Figure

How noise ratio and noise figure are defined?

Signal to Noise ratio (S/N) (信噪比)

$\dfrac{S}{N}=\dfrac{P_{signal}}{P_{noise}}\quad\text{(in ratio)}$

$\left(\dfrac{S}{N}\right)_{dB}=10\log\dfrac{P_{signal}}{P_{noise}}=10\log\dfrac{S}{N}$

Noise Ratio (NR，噪声比) and Noise Figure (NF，噪声因数)

$NR=\dfrac{S_i/N_i}{S_o/N_o}\quad\text{(in ratio)}$

$NF=(NR)_{dB}=10\log NR$

$10\log\dfrac{S_i/N_i}{S_o/N_o}=10\log(S_i/N_i)-10\log(S_o/N_o)$

$NR=\dfrac{SNR_1}{SNR_2}$

$NF=NR\;in\;dB=SNR_1-SNR_2\;in\;dB$

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Question 3

58

Bnad-pass Filter Parameters

In a band-pass filter, how the following terms are defined? Centre frequency, resonant frequency, cut-off frequency, bandwidth, quality factor.

Center frequency $\omega_0$ is defined as the frequency for which a circuit's transfer function is purely real, $\omega_0=\sqrt{\omega_{c1}\omega_{c2}}$.

Resonant frequency is the same as $\omega_0$, and when a circuit is driven at the resonant frequency, we say that the circuit is in resonance.

Cutoff frequency $\omega_{c1}$ and $\omega_{c2}$ are defined as the frequencies for which the magnitude of the transfer function equals $\left(\frac{1}{\sqrt{2}}\right)H_{max}$

Bandwidth $\beta$ is the width of the pass-band.

Quality factor Q (质量因数) is the ratio of the center frequency to the bandwidth $\frac{\omega_0}{\beta}$

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Question 4

59-76

Series/Parallel RLC Circuits

In series RLC and parallel RLC circuits, how can these terms mentioned in Q3 be derived?

Series RLC Circuit

1. Bandwidth $\beta=\omega_{c2}-\omega_{c1}=\frac{R}{L}$

2. Center frequency $\omega_0=\sqrt{\frac{1}{LC}}=\sqrt{\omega_{c1}\omega_{c2}}$

3. Quality factor $Q=\frac{\omega_0}{\beta}=\sqrt{\frac{L}{CR^2}}$

Parallel RLC Circuit

1. Bandwidth $\beta=\omega_{c2}-\omega_{c1}=\frac{1}{LC}$

2. Center frequency $\omega_0=\sqrt{\frac{1}{LC}}=\sqrt{\omega_{c1}\omega_{c2}}$

3. Quality factor $Q=\frac{\omega_0}{\beta}=\sqrt{\frac{CR^2}{L}}$

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Question 5

94-95

Pros and Cons of Negative Feedback System

What are the advantages and disadvantages of negative feedback system?

Advantages of Negative Feedback

1. Gain Sensitivity – variations in gain is reduced.

2. Bandwidth Extension – larger than that of basic amplified.

3. Noise Sensitivity – may increase S - N ratio.

4. Reduction of Nonlinear Distortion

5. Control of Impedance Levels – input and output impedances can be increased or decreased.

Disadvantages of Negative Feedback

1. Circuit Gain – reduced compared to that of basic amplifier.

2. Stability – possibility that feedback circuit will become unstable and oscillate at high frequencies.

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Question 6

103

Power Conversion Efficiency in Power Amplifiers

What is power conversion efficiency in Power Amplifiers?

Power conversion efficiency : $\eta=\dfrac{P_L}{P_S}$

where $P_L$ is signal load power, and $P_S$ is supply power

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Question 7

102-122

Typical Power Conversion Efficiency of Class ABC

What is the typical power conversion efficiency of Class A, Class B, Class C?

1. Class - A : an output transistor is biased at a quiescent current I Q and conducts for the entire cycle of the input signal
2. Class B : an output transistor conducts for only one - half of each sine wave input cycle
3. Class - AB : an output transistor is biased at a small quiescent current I Q and conducts for slightly more than half a cycle
4. Class - C : an output transistor conducts for less than half a cycle

Class A: $\eta(max)=25%$

Class B: $\eta(max)=78.5%$

Class C: larger than 78.5%

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Question 8

131-143

Use OP-amp to Construct the Mathematical Computation Circuits

How to use Op - amplifier to construct the mathematical computation circuits? Such as addition/subtraction, multiplication/division, log/anti - log, and etc.

None......

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Question 9

157

Barkhausen Criteria

What is Barkhausen criteria?

$$H(s)=\dfrac{X_o(s)}{X_i(s)}=\dfrac{A(s)}{1+A(s)\beta(s)}$$

Barkhausen Criteria

• $|A||\beta|=1$

• $\theta_A+\theta_B=2n\pi$ (where $n$ is an integer)

See also P185

this should be true only at $\omega_0$ (the frequency of oscillations), i.e. the feedback should be frequency selective

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Question 10

163

Wien Oscillator

Write down the feedback structure of a Wien Oscillator, and identify β.

Feedback structure

$\beta=\dfrac{V_{out}}{V_{in}}=\dfrac{Z_p}{Z_s+Z_p}$

$Z_s=R+\dfrac{1}{j\omega C}=\dfrac{1+\frac{j\omega}{\omega_1}}{j\omega C}$

$Z_p=\frac{\frac{R}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{1+j\omega RC}=\frac{R}{1+\frac{j\omega}{\omega_1}}$

Thus

$$\beta=\frac{1}{3}\angle 0$$

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Question 11

180

Shortage of Wien Oscillator

Why the Wien oscillator is not suitable for signal generation higher than 1 MHz?

$$\omega=2\pi f=\frac{1}{RC}\Rightarrow f=\frac{1}{2\pi RC}$$

$C\propto \frac{1}{f}$. For high frequency of oscillation $C$ will be small. When $f>1 MHz$, $C$ becomes comparable with the stray capacitance (杂散电容). For higher frequency operation, R must be reduced and this will increase the power dissipation. Hence the Wien oscillator is unattractive above 1 MHz. At higher frequencies, lower loss feedback networks are needed such as LC circuits or crystal resonators (晶体谐振器).

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Question 12

197-202

Circuit Transformation between Series and Parallel

How to perform the circuit transformation between series and parallel?

With $Q=X_s/R_s\gg 1$

$R_p\approx \frac{X_s^2}{R_s}$ and $X_p\approx X_s$, i.e. $L_p\approx L_s$

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Question 13

210-213

Colpitts Circuit

Perform small signal analysis method on Colpitts circuit, and find the oscillation criteria for the circuit.

None......

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Question 14

205 208 215

Resonant Frequency

What is the resonant frequency $f_r$ for Hartley, Colpitt, Clap circuits?

Hartley oscillator circuit
$f=\frac{1}{2\pi \sqrt{(L_1+L_2)\times C}}$
Colpitts oscillator circuit
$f=\frac{1}{2\pi \sqrt{C_1C_2/(C_1+C_2)\times L}}$
Clapp oscillator circuit
$f=\frac{1}{2\pi \sqrt{C_3L}}$

Clapp oscillator has better frequency stability than the other two oscillators.

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Question 15

223-226

Clapp Oscillator Circuit

For the Clapp oscillator circuit, how to performance design in terms of negative resistance?

Output frequency: $f=\dfrac{1}{2\pi\sqrt{C_3L}}$

Not finished yet......

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Question 16

251-253

Schmitt Trigger Oscillator

How to find the oscillation frequency of a Schmitt trigger oscillator?

$V(t)=V_1+(V_2-V_1)[1-\exp{\frac{t_1-t}{\tau}}]$
where $\tau=RC$
then
$T_1=RC\ln[\frac{V_{o+}-V_{i-}}{V_{o+}-V_{i+}}]$
$T_2=RC\ln[\frac{V_{i+}-V_{o-}}{V_{i-}-V_{o-}}]$
whence the frequency of oscillation is: $f=\dfrac{1}{T_1+T_2}$

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Question 17

263-265

Phase Detection Achieved by Multiplier

How Phase detection can be achieved by a multiplier?

We often want a circuit that will give an output that is proportional to the difference of phase of two equal frequency sinusoids.

For example,

$V_1=m(t)\cos(\omega_ct+\theta_1)$
$V_2=m(t)\cos(\omega_ct+\theta_2)$
$V_o=K_D(\theta_1-\theta_2)$

Answer:

By using a phase detector consists of a ultiplier and a low-pass filter (LPF):
$V_o=cV_1V_2\cos(\omega_ct+\theta_1)\cos(\omega_ct+\theta_2)$
$V_o=\frac{1}{2}cV_1V_2(\cos(\theta_1-\theta_2)+\cos(2\omega_ct+\theta_1+\theta2))$
$V_o=\frac{1}{2}cV_1V_2(\cos(\theta_1-\theta_2)$
The right part is removed by LPF

Let $\theta_1-\theta_2=\pi/2+\phi$ then
$V_{out}=\frac{1}{2}cV_1V_2\cos(\pi/2+\phi)=-\frac{1}{2}cV_1V_2\sin\phi$

If $\phi$ is small then $V_{out}\approx-(1/2)cV_1V_2\phi$

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Question 18

270-272

Analogue Multiplier Circuit

Identify the analogue multiplier circuit, and derive the equation for output voltage.

Assume Q1 and Q2 are well matched (made on the same chip for equality of characteristics and temperatures)

Q1 and Q2 are differential amplifier, $V_{out}=V_{o1}-V_{o2}$

Step 1

With $V_2=0$, Q1 and Q2 will both have quiescent (静止的) current:
$I_1=I_2=I/2$
and hence have mutual conductance:

$$g_{m1}=g_{m2}=g_m=\frac{q}{k_BT}\frac{I}{2}=\frac{1}{V_T}\frac{I}{2}=40\frac{I}{2}$$

where $V_T=k_BT/q\approx 25\text{ mV}$, $q$ is the electron charge, $k_B$ is the Boltzmann constant, $T$ is the absolute temperature.

Step 2

With $V_2\neq 0$,
$V_{o1}=V_{cc}-I_1R_c$, $V_{o2}=V_{cc}-I_2R_c$
$V_{out}=V_{o1}-V_{o2}=(I_2-I_1)R_c$

When $V_2$ is small (otherwise, this will have exponential behavior), we have:

$I_1=\frac{I}{2}+g_m\frac{V_2}{2}$
$I_2=\frac{I}{2}-g_m\frac{V_2}{2}$
$\therefore I_2-I_1=-g_mV_2$

Step 3

Whence
$V_{out}=V_{o1}-V_{o2}=(I_2-I_1)R_c$
$I_2-I_1=-g_mV_2$
$g_m=\frac{1}{V_T}\frac{I}{2}$
$\therefore V_{out}=-\dfrac{R_cIV_2}{2V_T}$

Recall:
For triode, $I_C=\frac{\beta}{\beta+1}I_E$
$\beta$ is often greater than 100, then $I_C\approx I_E$

Step 4

$V_3\approx V_1$
$I_3=V_3/R_e=V_1/R_e$
$I\approx I_3$
$\therefore I\approx \dfrac{V_1}{R_e}$

Step 5

$V_{out}=-\dfrac{R_c}{2V_TR_e}V_1V_2=kV_1V_2$
$k=-\dfrac{R_c}{2V_TR_e}$

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Question 19

274-277

Four-quafrant Multiplier Circuit

Identify the four-quadrant multiplier circuit, and derive the equation for output voltage.

Step 1

$V_{o2}=V_{cc+}-I_{o2}R_c$

With the assumption Q1=Q2, Q3=Q4=Q5=Q6, and with $V_1=V_2=0$

$I_1=I_2=2I_o$, $I_{11}=I_{12}=I_{21}=I_{22}=I_o$
$V_{o2}=V_{cc+}-I_{o2}R_c=V_{cc+}-2I_oR_c=V_{o1}$ ($I_{o2}=I_{12}+I_{22}$)

For Q1 and Q2, $g_m=I_c/V_T=2I_o/V_T$

For Q3 and Q4, $g_{m1}=\frac{I_1}{2V_T}$
For Q5 and Q6, $g_{m2}=\frac{I_2}{2V_T}$

$g_{m2}-g_{m1}=\frac{I_2-I_1}{2V_T}$

Step 2

When $V_1\neq 0$

$I_1=2I_o+g_mV_1/2$
$I_2=2I_o-g_mV_1/2$

$I_2+I_1=4I_o$
$I_2-I_1=-g_mV_1$

When $V_2\neq 0$

$I_{o2}=I_{12}+I_{22}$
$I_{12}=I_1/2-g_{m1}V_2/2$
$I_{22}=I_2/2+g_{m2}V_2/2$
（注意图中$V_2$在外围两侧，而不是两个差分放大器的各自左右侧）

$\begin{aligned} I_{o2}&=I_{12}+I_{22}\\ &=(I_1+I_2)/2+V_2(g_{m2}-g_{m1})/2\\ &=2I_o+\frac{V_2}{2V_T}(I_2-I_1)/2\\ &=2I_o+\frac{V_2}{2V_T}(-g_mV_1)/2\\ &=2I_o-\frac{1}{2V_T}\frac{2I_o}{V_T}\frac{V_1V_2}{2} \end{aligned}$

With $c_1=\frac{I_o}{2V_T^2}$, $I_{o2}=2I_o-c_1V_1V_2$

Step 3

$V_{o2}=V_{cc+}-I_{o2}R_c=V_{cc+}-(2I_o-c_1V_1V_2)R_c=(V_{cc+}-2I_oR_c)+c_2V_1V_2$
where $c_2=\frac{I_oR_c}{2V_T^2}$, $V_{cc+}-2I_oR_c$ is called quiescent voltage.

Take a "double ended" output:

$V_{o1}=(V_{cc+}-2I_oR_c)-c_2V_1V_2$

$$V_o=V_{o2}-V_{o1}=2c_2V_1V_2=\frac{I_oR}{V_T^2}V_1V_2$$

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Question 20

283-285

Signal Cycle of a Ring Modulator

Explain the signal cycle of a ring modulator.

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Question 21

313-315

CB Circuit vs. CE Circuit

Explain why a CB circuit usually has a better performance for high frequency signal if compared with a CE circuit

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Question 22

325-326

Expanded Bandwidth by Feedback

Explain how bandwidth could be expanded by feedback.

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Question 23

349-352

Phase-locked Loop Circuit

Given a phase-locked loop circuit, perform the in-lock transient analysis, and find the transfer function $T(s)=\frac{\theta_{out}}{\theta_{in}}$.Then, identify $\omega_n$ and $\xi$ for the second order system.

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Question 24

370-371

Possible States of Phase Locked Loop System

What are the possible states of operation of a phase locked loop system?

1. Free running
2. Capture
3. Locked or tracking