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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?

Jason R
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Cell-o
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    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? $$H(s) = \frac{s}{(s+1)(s+2)} = \frac{a}{s+1}+\frac{b}{s+2}$$ – Dilip Sarwate Dec 22 '11 at 21:50
  • Firstly,thanks for your interesting.As you mentioned,I applied partial fractions.I find a and b in the equation.(a=-1,b=2). – Cell-o Dec 22 '11 at 22:03
  • 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? – Dilip Sarwate Dec 22 '11 at 22:05
  • 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. – Cell-o Dec 22 '11 at 22:15
  • @DilipSarwate At this level it's safe to assume that they teach them BIBO stability. – Phonon Dec 22 '11 at 22:52
  • Any suggestions? – Cell-o Dec 23 '11 at 09:10

1 Answers1

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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.

Dipan Mehta
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  • thanks so much.I want to ask a last question.How can I find the gain and phase response of the above system? – Cell-o Dec 24 '11 at 21:47
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    -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." – Dilip Sarwate Dec 24 '11 at 22:00
  • 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. – Dilip Sarwate Dec 25 '11 at 19:22
  • @DilipSarwate I was surprised to know that ideal LPF can be or is BIBO unstable. I am following the definition here from wiki- http://en.wikipedia.org/wiki/BIBO_stability. Can you give me proof? – Dipan Mehta Dec 26 '11 at 15:44
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    @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. – Dipan Mehta Dec 26 '11 at 15:47
  • "I don't have any apprehension to know if the math needs corrections" Please don't take the square root of a negative number as in $-1\cdot \omega^2$ no matter how long you are out of school. I cannot revise the question since I don't have the necessary reputation and while I can certainly write out an answer, I don't like to spoon-feed answers to homework; students need to learn these things on their own. What is bad is giving them totally incorrect information in the guise of an answer. What is worse is when the poor unsuspecting student accepts this mis-information as the answer. – Dilip Sarwate Dec 26 '11 at 16:47
  • @DilipSarwate Ok. About SQRT of (-1): The term $-1\cdot\omega^2$ arrived by as $j^2 = -1$. To avoid taking the square root again, that 20 Log (x) would take care of it. I didn't knew if there is a typo. Anyway, for me this is learning so i requested you. – Dipan Mehta Dec 26 '11 at 17:20
  • Since you refuse to consider the possibility of a mistake in your answer, please apply your formula and tell us the magnitude of $H(j\omega)$ at $\omega = 1$. Remember that whether you take the square root or not, or the logarithm or not, there is a term $4 - 5\omega^2 + \omega^4$ in the denominator. Remember that according to you, the frequency response of a BIBO-stable system needs to be finite at all frequencies (which should include $\omega = 1$. Is the system BIBO-stable? – Dilip Sarwate Dec 27 '11 at 13:27
  • I corrected the stability definition after all inputs. About the $\omega=1$ condition, yes, this is an interesting point. I am never denying any mistake. Since you have clearly told there is a mistake - i believe the correction is required. If you can put forth your suggestion, will move towards the right answer or else at least give direction so i will put the effort to fix. I want to see that finally, we leave the answer in a fit state and i learn what is missing. – Dipan Mehta Dec 27 '11 at 15:43
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    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? – Dilip Sarwate Dec 27 '11 at 22:20