# Determining the causality of a signal with it's pole-zero plot

I have the following question:

Pole-zero plot of x(t) and y(t) are given below: The signal $$g(t)$$ and $$h(t)$$ are defined as $$g(t)=x(t)e^{-3t}$$ and $$h(t)=y(t)*e^{-t}u(t)$$. If $$g(t)$$ and $$h(t)$$ are both absolutely integrable, determine whether the signals $$g(t)$$, $$h(t)$$ are left-sided/right-sided.

My try:

I take the laplace transform of both the signals and get $$G(s)=X(s+3)$$ and $$H(s)=Y(s)\cdot \frac{1}{s+1}$$

Also because both $$g(t)$$ and $$h(t)$$ are absolutely integrable, their transforms must be stable so both must have $$jw$$-axis in their respective ROC.

As we can see $$X(s+3)$$ shifts the pole-zero plot to the left by $$3$$ units so we have all the poles in the left $$s$$-plane and ROC would be $$Re\{s\}>-1$$ hence $$g(t)$$ is right sided.

Similarly $$H(s)$$ has all the poles in the left $$s$$-plane and ROC is again $$Re\{s\}>-1$$ hence $$h(t)$$ is right sided.

Is this reasoning correct?

• yes it's a correct reasoning... – Fat32 Oct 28 '18 at 11:06
• @Fat32 Thanks for the confirmation. You may use it as an answer with some extra points (if required). – paulplusx Oct 28 '18 at 11:16
• No thanks. A comment seems enough here... – Fat32 Oct 28 '18 at 11:19
• @Fat32 But then this question would be "unanswered" as per SE system, right? Is that ok? – paulplusx Oct 28 '18 at 11:21
• yes that's right it will be unanswered. So let's make one ;-) – Fat32 Oct 28 '18 at 11:23

Looking at the pole-zero plots of the continuous-time signals $$x(t)$$ and $$y(t)$$, and the new signals $$g(t)= x(t)e^{-3t}$$ and $$h(t) = y(t) \star e^{-t}u(t)$$, the pole locaitons of $$g(t)$$ and $$h(t)$$ are found to be:

$$g(t) = x(t)e^{-3t} \implies G(s) = X(s+3) \implies \text{ Re(poles) } = \{-1,-1 \}$$

$$h(t) = y(t) \star e^{-t}u(t) \implies H(s) = Y(S) \frac{1}{s+1} \implies \text{ Re(poles) } = \{-2,-2,-1 \}$$

From these pole locations we see the following:

1-$$g(t)$$ has two possible ROCs: $$Re\{s\} <-1$$, and $$Re\{s\} > -1$$, and only the second one includes the $$j\omega$$ axis and hence can be stable.

2-$$h(t)$$ has three possible ROCs: $$Re\{s\}<-2$$, $$-2 < Re\{s\} <-1$$, and $$Re\{s\} > -1$$, and only the last one includes the $$j\omega$$ axis and is stable.

So for $$h(t)$$ and $$g(t)$$ to be absolutely integrable (stable), their ROC's must include $$j\omega$$ axis and this means their ROCs are to the right of the largest poles which implies that the signals are causal right sided signals.

• Yes, you included those extra points because the question is so vague. Thanks a lot :-) – paulplusx Oct 28 '18 at 13:33
• your welcome @paulplusx ... as a jack of all trades what's your main training focus ? – Fat32 Oct 28 '18 at 13:36
• Still struggling to find :P I like electronics and programming and how beautifully they come together :-) Still an aspiring masters student though :P – paulplusx Oct 28 '18 at 14:13
• Yes that's the case; electronics and programming stem from the same branch and complement each other very beautifully. That's why there are departments like Electrical and Computer Engineering (ECE) around the world. – Fat32 Oct 28 '18 at 14:35
• Yes, very true, I am aiming for that particular department ECE (Computer engineering to be specific). Unfortunately and sadly the country I am in doesn't have that ECE. Here, electrical, electronics and computer science are separate departments. Here, ECE means Electronics and Communication Engineering :-( which was there in my bachelor's. Still not a real problem, knowledge is everywhere :-) – paulplusx Oct 28 '18 at 15:04