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I have the following transfer function:

$$H(s)=\frac{s}{(s+1)(s+2)}$$

How can I find the gain and phase response of the above system? I know the first step has something to do with substituting $s = j\omega$ into $H(s)$. How can I find whether the system is stable and whether it is causal?

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    $\begingroup$ What is the definition of gain of a system, or stability of a system that you have been taught? Can you look at $H(s)$ and apply the condition directly? Do you understand the notion of partial fractions? or if not, can you solve for both $a$ and $b$ in the following equation? $$H(s) = \frac{s}{(s+1)(s+2)} = \frac{a}{s+1}+\frac{b}{s+2}$$ $\endgroup$
    – Dilip Sarwate
    Dec 22, 2011 at 21:50
  • $\begingroup$ Firstly,thanks for your interesting.As you mentioned,I applied partial fractions.I find a and b in the equation.(a=-1,b=2). $\endgroup$
    – Cell-o
    Dec 22, 2011 at 22:03
  • $\begingroup$ Good. Now can you look in your Laplace transform table and figure out that $h(t)$, the inverse Laplace transform of $H(s)$, is the sum of two decaying exponential functions? $\endgroup$
    – Dilip Sarwate
    Dec 22, 2011 at 22:05
  • $\begingroup$ I find that. $H(t)=-{{e}^{-t}}u(t)+2{{e}^{-2t}}u(t)$ is that correct?Because,I look transform pairs in Laplace transform table.There are two ROC.So $\operatorname{Re}(s)>-\operatorname{Re}(a)$ and $\operatorname{Re}(s)<-\operatorname{Re}(a)$. Which of them will be used? Thanks. $\endgroup$
    – Cell-o
    Dec 22, 2011 at 22:15
  • $\begingroup$ @DilipSarwate At this level it's safe to assume that they teach them BIBO stability. $\endgroup$
    – Phonon
    Dec 22, 2011 at 22:52

1 Answer 1

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NOTE: This answer is community wiki: If you feel there is a need for correction -please update the answer right here.

1. Frequency response:
For any system - the (discrete time or continuous time) Fourier Transform of the impulse response is same as the frequency response of it. You can replace $s$ by $j\omega$ to work on this.

2. Stability:
System is stable if, the bounded input produces bounded output. The definition of can be found here in the wiki link. For this, the necessary condition to prove is to see if the impulse response be absolutely integrable, i.e., its L^1 norm exist.

For a rational and continuous-time system, the condition for stability is that the region of convergence (ROC) of the Laplace transform includes the imaginary axis.

3. Causality:
Any system who's impulse response at sample $s_i$ doesn't require to know any samples presented $s_{i+1}$ or after wards, than the system is causal.

EDIT:
Now given the above i am trying to put the calculation as per your system.

$$ H(s)=\frac{s}{(s+1)(s+2)} $$

Now we can replace $s$ with $j\omega$ to get the frequency response in terms of magnitude and phase response.

$$ H(j\omega) = \frac{j\omega}{(j\omega+1)(j\omega+2)} $$ $$ H(j\omega) = \frac{j\omega}{2 - \omega^2 + 3j\omega } $$ $$ Magnitude = 20 Log_{10} \sqrt { ( H(j\omega) )^2 } $$ $$ Magnitude = 20 Log_{10} \sqrt { \frac { (j\omega)^2 }{(2 - \omega^2)^2 + (3j\omega)^2} } $$

NOTE: square is applied separately on real terms than on imaginary terms! I got this understanding from this document here.

Replacing $-j^2$ by $-1$

$$ Magnitude = 20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(2 - \omega^2)^2 - (3\omega)^2} } $$

$$ Magnitude = 20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(4 - 5\omega^2 +\omega^4)} } $$

$$ Phase = arc tan (real / imaginary) $$

$$ Phase = -1 \cdot \tan^{-1}(\frac {-3\omega}{(2-\omega^2)} ) \ $$

Please correct me if i am wrong.

Another literature that can help you.

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  • $\begingroup$ thanks so much.I want to ask a last question.How can I find the gain and phase response of the above system? $\endgroup$
    – Cell-o
    Dec 24, 2011 at 21:47
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    $\begingroup$ -1 for a poorly thought-through response! (i) What is meant by the sample space that is $s$? (ii) Since you have not defined what you mean by gain, I wonder what it means to write "gain is finit (sic) and not changing over time"? Since at least in part 3 you are discussing a discrete time system, is the gain of a discrete-time system at time $i$ the value of the impulse response at time $i$? If not, what is it? (iii) a voltage divider has frequency response $H(\omega) = 0.5$ for all $\omega$. According to 2., it is stable (finite gain) and unstable also: $H(\omega)$ does not "go to zero." $\endgroup$
    – Dilip Sarwate
    Dec 24, 2011 at 22:00
  • $\begingroup$ While the editing has improved the answer a little, there are still many problems that are unresolved. For example, the definition of stability is sort of but not exactly that of BIBO stability mentioned by phonon, but it is not true that if the frequency response is bounded, the system is BIBO stable. An ideal low-pass filter is an example of a system that is not BIBO stable even though its frequency response is bounded for all $f$. Most important, this accepted answer does not address any of the questions raised by the OP; it merely supplies definitions that the OP ought to know. $\endgroup$
    – Dilip Sarwate
    Dec 25, 2011 at 19:22
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    $\begingroup$ @DilipSarwate I don't have any apprehension to know if the math needs corrections or improvement. It's been some good 11 years since i left school! i would appreciate if you kindly revise the question - or at least write your answer and for sure i would sure upvote it. $\endgroup$
    – Dipan Mehta
    Dec 26, 2011 at 15:47
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    $\begingroup$ A causal system with impulse response $h(t)=e^{-t}$ for $t>0$ has absolutely integrable impulse response, and is thus BIBO-stable. All bounded inputs produce bounded outputs. The frequency response is $$H(j\omega)=\int_{-\infty}^\infty h(t)e^{-j\omega t}dt=\int_0^\infty e^{-t}e^{-j\omega t}dt=\frac{1}{1+j\omega}$$ and so a unit-amplitude sinusoidal input at frequency $\omega$ rad/sec produces sinusoidal output of amplitude $$|H(j\omega)|=\frac{1}{1^2+(j\omega)^2}=\frac{1}{1-\omega^2}$$ per your formula. So, if $\omega=1$, bounded input produces unbounded output. Could something be wrong? $\endgroup$
    – Dilip Sarwate
    Dec 27, 2011 at 22:20

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