Electrical Engineering: Is this signal stable?

  • #1
Boltzman Oscillation
233
26

Homework Statement



y(t) = x(1-5t)

Homework Equations



none


The Attempt at a Solution



well I've never looked at the stability of a signal which has a time scale and shift. My guess is that it is stable as anything I can provide as input will output a bounded signal.

Ex: if x(t) is u(t)
then y(t) = u(1-5t) which is bounded.
Am I correct or no? Also can I prove that it is stable just by one example or do i have to find a way to generalize?
 

Answers and Replies

  • #2
DaveE
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I don't think you've provided enough information to get useful comments; at least not by me.
What is x(t)? What is u(t) (the step function, maybe)?
What if x(t) = 1 - t? then y(t) = 5t
 
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  • #3
Boltzman Oscillation
233
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I don't think you've provided enough information to get useful comments; at least not by me.
What is x(t)? What is u(t) (the step function, maybe)?
What if x(t) = 1 - t? then y(t) = 5t

Sorry, yes the u(t) is the step function.
Im given the question:
Determine whether the following system is stable:

y(t) = x(1 - 5t)

this is for a signals class in my electrical engineering corriculumn.
 
  • #4
DaveE
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Sorry, yes the u(t) is the step function.
Im given the question:
Determine whether the following system is stable:

y(t) = x(1 - 5t)

this is for a signals class in my electrical engineering corriculumn.
And yet we still know nothing about the function x(t) in your problem statement.
With no restrictions on x(t), I can pretty much guarantee that there is a function x(t) for which x(1-5t) is unstable; there are also stable ones.
If you want a useful answer, you must ask a complete question.
 
  • #5
jasonRF
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What is the definition of "stable" that you are using? I have a guess based on my own EE background, and I am guessing once you write out the definition it will be clear to you whether this is stable or unstable.
 
  • #6
Boltzman Oscillation
233
26
What is the definition of "stable" that you are using? I have a guess based on my own EE background, and I am guessing once you write out the definition it will be clear to you whether this is stable or unstable.
From my own words, a stable system is one where a bounded input will create a bounded output at all times.
 
  • #7
AVBs2Systems
39
24
Hi

I think the title of this thread should be is this system stable instead of is this signal stable, because ## \textbf{BIBO stability} ## is a system property and not a signal property. There are a few system properties:
1. BIBO stability.
2. Causality.
3. Linearity.
4. Time invariance.
5. Memory (with or without).
6. Invertibility.

A system is ## \textbf{T} ## is a mathematical relation between an input and an output signal. We assume single variable functions of time here:
$$
\textbf{T} \{ x(t) \} = y(t)
$$
In the frequency domain where ## H(s) ## is the transfer function:
$$
H(s) \cdot X(s) = Y(s)
$$

A system is BIBO stable if:
1. For every bounded input, it produces a bounded output. AND
2. Its impulse response ## h(t) = \textbf{T} \{\delta(t) \} ## is absolutely integrable (the impulse response is bounded and an energy signal) . AND
3. All of its poles of the transfer function have a strictly negative real part.
4. Possibly other deeper requirements.

If you can prove one of the above requirements (any of 1 to 3) you have proven the system is bibo stable.

This system is described with:

$$
\textbf{T}\{x(t) \} = x(-5t + 1)
$$
The impulse response is:
$$
h(t) = \textbf{T}\{ \delta(t) \} = \dfrac{1}{5} \cdot \delta(t + 1)
$$
The above function is absolutely integrable, but the energy and power of this function cannot be defined. So, it is a bibo stable system.
Intuitively, just look at what the system is doing:
It flips, compresses, and time shifts your signals None of these operations can remove the boundedness from a bounded function or signal.
 
Last edited:
  • #8
Boltzman Oscillation
233
26
Hi

I think the title of this thread should be is this system stable instead of is this signal stable, because ## \textbf{BIBO stability} ## is a system property and not a signal property. There are a few system properties:
1. BIBO stability.
2. Causality.
3. Linearity.
4. Time invariance.
5. Memory (with or without).
6. Invertibility.

A system is ## \textbf{T} ## is a mathematical relation between an input and an output signal. We assume single variable functions of time here:
$$
\textbf{T} \{ x(t) \} = y(t)
$$
In the frequency domain where ## H(s) ## is the transfer function:
$$
H(s) \cdot X(s) = Y(s)
$$

A system is BIBO stable if:
1. For every bounded input, it produces a bounded output. AND
2. Its impulse response ## h(t) = \textbf{T} \{\delta(t) \} ## is absolutely integrable (the impulse response is bounded and an energy signal) . AND
3. All of its poles of the transfer function have a strictly negative real part.
4. Possibly other deeper requirements.

If you can prove one of the above requirements (any of 1 to 3) you have proven the system is bibo stable.

This system is described with:

$$
\textbf{T}\{x(t) \} = x(-5t + 1)
$$
The impulse response is:
$$
h(t) = \textbf{T}\{ \delta(t) \} = \dfrac{1}{5} \cdot \delta(t + 1)
$$
The above function is absolutely integrable, but the energy and power of this function cannot be defined. So, it is a bibo stable system.
Intuitively, just look at what the system is doing:
It flips, compresses, and time shifts your signals None of these operations can remove the boundedness from a bounded function or signal.
I see. Thank you very much. I learned some new terminology and a more general way of determining stability. Thanks again.
 
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