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Chapter 7: Problem 9

(a) (i) Sketch the Bode magnitude plot for the function $$ T(s)=\frac{10(s+10)(s+100)}{(s+1)(s+1000)} $$ (ii) What are the corner frequencies? (iii) Determine \(|T(\omega)|\) for \(\omega \rightarrow 0\). (iv) Determine \(|T(\omega)|\) for \(\omega \rightarrow \infty\). (b) Repeat part (a) for the function $$ T(s)=\frac{8 s^{2}}{(0.2 s+1)^{2}} $$

Short Answer

Expert verified
(i) Sketch each step based on the identified corner frequencies. (ii) 1, 10, 100, 1000 rad/s and 5 rad/s. (iii) Function 1: 20 dB; Function 2: 18 dB. (iv) Function 1: 20 dB; Function 2: 0 dB.

Step by step solution

01

Identify Corner Frequencies for Function 1

The function is given by \[ T(s)=\frac{10(s+10)(s+100)}{(s+1)(s+1000)} \]. Identify the corner frequencies by finding the poles and zeros:- Zeros at \( s = -10 \) and \( s = -100 \), giving corner frequencies of 10 rad/s and 100 rad/s.- Poles at \( s = -1 \) and \( s = -1000 \), giving corner frequencies of 1 rad/s and 1000 rad/s.
02

Sketch Bode Magnitude Plot for Function 1

Construct the Bode plot by considering each zero and pole:- Start at a gain of 20 dB due to the DC gain factor (10), equivalent to 20 \( \log_{10}(10) \).- At 1 rad/s, a pole contributes -20 dB/decade (slope decrease).- At 10 rad/s, a zero adds +20 dB/decade (slope increase).- At 100 rad/s, a zero adds another +20 dB/decade.- At 1000 rad/s, a pole adds -20 dB/decade.Consider changes in slope at each corner frequency.
03

Determine |T(\omega)| as \(\omega \rightarrow 0\) for Function 1

For \( \omega \rightarrow 0 \), the gain is dominated by the constant terms and low-frequency approximation:\[ |T(0)| = \frac{10 \times 10 \times 100}{1 \times 1000} = 10. \]In dB, this equates to 20 dB.
04

Determine |T(\omega)| as \(\omega \rightarrow \infty\) for Function 1

For \( \omega \rightarrow \infty \), as the powers of \( s \) in the numerator and denominator are equal, the system reaches a constant gain at high frequencies.Calculate using coefficients of the highest powers of \( s \):\[ |T(\infty)| = \frac{10 \times 1 \times 1}{1 \times 1} = 10, \]equivalent to 20 dB.
05

Identify Corner Frequencies for Function 2

The function is given by \[ T(s)=\frac{8s^{2}}{(0.2s+1)^{2}} \]. Identify the corner frequencies:- Zeros: \( s = 0 \rightarrow \omega = 0 \) rad/s,- Poles: two identical poles at \( s = -5 \) from \( 0.2s+1 = 0 \rightarrow s = -5 \), giving a corner frequency of 5 rad/s.
06

Sketch Bode Magnitude Plot for Function 2

Construct the Bode plot:- Start at 18 dB due to DC gain factor (8), \( 20\log_{10}(8) = 18.06 \) dB. Simplified to 18 dB.- Two zeros at the origin contribute +40 dB/decade starting from \( \omega = 1 \) rad/s.- Double poles at 5 rad/s contribute -40 dB/decade starting from 5 rad/s.Summarize the contributions at each frequency.
07

Determine |T(\omega)| as \(\omega \rightarrow 0\) for Function 2

For \( \omega \rightarrow 0 \), the gain is determined by the zero at the origin, i.e., theoretically, the gain approaches infinity, but practically, focusing on low \( \omega \), it starts at 18 dB.
08

Determine |T(\omega)| as \(\omega \rightarrow \infty\) for Function 2

For \( \omega \rightarrow \infty \), as the zeros and poles cancel out, leaving the distributed gain associated with high frequencies:\[ |T(\infty)| = \frac{1}{1} = 1, \]equivalent to 0 dB.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Corner Frequencies
In Bode plot analysis, corner frequencies are crucial marker points on the plot. They determine where the slope of the graph changes due to the presence of poles and zeros. A pole results in a downward slope, while a zero results in an upward slope. For example, in the function \( T(s)=\frac{10(s+10)(s+100)}{(s+1)(s+1000)} \), the corner frequencies are determined by the values where these poles and zeros occur.
The zeros at \( s = -10 \) and \( s = -100 \) create corner frequencies at 10 rad/s and 100 rad/s, respectively. These will cause the slope of the Bode plot to rise. On the other hand, the poles at \( s = -1 \) and \( s = -1000 \) establish corner frequencies at 1 rad/s and 1000 rad/s, respectively, causing the slope to fall.
  • Each zero increases the slope by +20 dB/decade.
  • Each pole decreases the slope by -20 dB/decade.
Identifying these frequencies helps to accurately sketch the Bode plot by knowing where the behavior of the plot changes.
High-Frequency Gain
The high-frequency gain in Bode plot analysis represents the gain of the system as \( \omega \rightarrow \infty \). This is where the behavior of the system is dominated by its highest-order terms. For the function \( T(s)=\frac{10(s+10)(s+100)}{(s+1)(s+1000)} \), the high-frequency gain can be determined by evaluating the ratio of the coefficients of the leading terms.
Here, both the numerator and denominator have quadratic leading terms, so the high-frequency gain is computed as \( \frac{10 \cdot 1 \cdot 1}{1 \cdot 1} = 10 \), which corresponds to 20 dB on the plot.
As high frequencies are approached, this constant slope stabilizes, showing how the system behaves when driven by high-frequency inputs. The key takeaway is that the gain 'flattens out', indicating a limiting value for the system's response at high frequencies.
Low-Frequency Gain
Low-frequency gain refers to the system's behavior as \( \omega \rightarrow 0 \). At low frequencies, the system behavior is largely defined by its constant and lower-order terms. For the function \( T(s)=\frac{10(s+10)(s+100)}{(s+1)(s+1000)} \), we can determine the low-frequency gain considering the direct current (DC) gain, which appears as the constant terms multiply and simplify.
At \( \omega = 0 \), the effects of the zeros and poles simplify to \( |T(0)| = \frac{10 \times 10 \times 100}{1 \times 1000} = 10 \). This can be represented as 20 dB on the Bode plot.
  • DC gain provides the starting point for the Bode plot.
  • The initial gain value sets the baseline for the plot before any changes in slope at corner frequencies take place.
Understanding low-frequency gain is essential for predicting how the system reacts to slow-moving signals or inputs with minimal frequency.

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Most popular questions from this chapter

Design a bipolar amplifier with a midband gain of \(\left|A_{v}\right|=50\) and a lower \(3 \mathrm{~dB}\) frequency of \(10 \mathrm{~Hz}\). The available transistors are \(2 \mathrm{~N} 2222 \mathrm{~A}\), and the available power supplies are \(\pm 10 \mathrm{~V}\). All transistors in the circuit should be biased at approximately \(0.5 \mathrm{~mA}\). The load resistance is \(R_{L}=5 \mathrm{k} \Omega\), the source resistance is \(R_{S}=0.1 \mathrm{k} \Omega\), and the input and output are ac coupled to the amplifier. Compare the bandwidth of a single-stage design to that of a cascode design.

A bipolar transistor has \(f_{T}=4 \mathrm{GHz}, \beta_{o}=120\), and \(C_{\mu}=0.08 \mathrm{pF}\) when operated at \(I_{C Q}=0.25 \mathrm{~mA}\). Determine \(g_{m}, f_{\beta}\), and \(C_{\pi}\).

The parameters of an ideal n-channel MOSFET are \(W / L=8\), \(\mu_{n}=400 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}, C_{\mathrm{ox}}=6.9 \times 10^{-7} \mathrm{~F} / \mathrm{cm}^{2}\), and \(V_{T N}=0.4 \mathrm{~V}\). (a) Determine the maximum source resistance such that the transconductance \(g_{m}\) is reduced by no more that 20 percent from its ideal value when \(V_{G S}=3 \mathrm{~V}\). (b) Using the results of part (a), find how much \(g_{m}\) is reduced from its ideal

(a) Design a common-emitter amplifier using a 2N2222A transistor biased at \(I_{C Q}=1 \mathrm{~mA}\) and \(V_{C E Q}=10 \mathrm{~V}\). The available power supplies are \(\pm 15 \mathrm{~V}\), the load resistance is \(R_{L}=20 \mathrm{k} \Omega\), the source resistance is \(R_{S}=0.5 \mathrm{k} \Omega\), the input and output are ac coupled to the amplifier, and the lower \(3 \mathrm{~dB}\) frequency is to be less than \(10 \mathrm{~Hz}\). Design the circuit to maximize the midband gain. What is the upper \(3 \mathrm{~dB}\) frequency? (b) Repeat the design for \(I_{C Q}=\) \(50 \mu \mathrm{A}\). Assume \(f_{T}\) is the same as the case when \(I_{C Q}=1 \mathrm{~mA}\). Compare the midband gain and bandwidth of the two designs.

(a) The frequency \(f_{T}\) of a bipolar transistor is found to be \(540 \mathrm{MHz}\) when biased at \(I_{C Q}=0.2 \mathrm{~mA}\). The transistor parameters are \(C_{\mu}=0.4 \mathrm{pF}\) and \(\beta_{o}=120\). Determine \(f_{\beta}\) and \(C_{\pi}\). (b) Using the results of part (a), determine \(f_{T}\) and \(f_{\beta}\) when the transistor is biased at \(I_{C Q}=0.8 \mathrm{~mA}\).
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