# Electrical Engineering: Is this signal stable?

Boltzman Oscillation

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

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?

Gold Member
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

• scottdave and berkeman
Boltzman Oscillation
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.

Gold Member
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.

Gold Member
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.

Boltzman Oscillation
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.

AVBs2Systems
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:
Boltzman Oscillation
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.

• AVBs2Systems