Necessary Condition on BIBO (Bounded Input Bounded Output) Stability In Sect. 1.2.8 it was shown that the absolute integrability/summability (1.2.27a)/(1.2.27b) of impulse response guarantees the BIBO stability and thus are sufficient conditions. Here, you can show that if the conditions do not hold, there is a bounded input yielding unbounded output, which implies that they are also necessary conditions for BIBO stability. To this end, suppose that the impulse response of a system does not satisfy the condition (1.2.27a)/(1.2.27b):
Are these signals bounded? If they are bounded, what is their upper bound?
(b) Show that the output y(t)/y[n] to the (bounded) input x(n)/x[n] at t = 0/n = 0 is unbounded, implying that the system is not BIBO stable
The Correct Answer and Explanation is :
To address the question, let’s first break down the concepts and the steps involved in demonstrating the necessary condition for Bounded Input Bounded Output (BIBO) stability:
Part (a) – Are the signals bounded?
We are given two impulse responses that are described in the form of functions or sequences, potentially involving either continuous-time signals (for ( t )) or discrete-time signals (for ( n )). The impulse response ( h(t) ) or ( h[n] ) typically characterizes a system’s response to a unit impulse input.
In the context of BIBO stability, the system is stable if, for any bounded input, the output remains bounded. The necessary condition for this stability is that the impulse response must be absolutely integrable in continuous time (i.e., ( \int |h(t)| \, dt < \infty )) or summable in discrete time (i.e., ( \sum |h[n]| < \infty )).
To determine if the given impulse responses are bounded, we must evaluate the signals. Based on the image provided (which you mentioned, but I cannot directly access), we assume the signals are either integrable or summable based on the given conditions. If the conditions for absolute integrability or summability do not hold, the signals would not be bounded.
Part (b) – Unbounded output and BIBO instability
Now, we need to show that when the impulse response does not satisfy the absolute integrability (continuous) or summability (discrete) condition, there exists a bounded input that produces an unbounded output, indicating that the system is not BIBO stable.
Let’s analyze this through a counterexample:
- Assumption: The impulse response ( h(t) ) (continuous-time) or ( h[n] ) (discrete-time) does not satisfy the condition for absolute integrability or summability. This means that:
- In continuous-time, ( \int |h(t)| \, dt = \infty ).
- In discrete-time, ( \sum |h[n]| = \infty ).
- Bounded input: Assume an input signal ( x(t) ) (continuous-time) or ( x[n] ) (discrete-time) is bounded, meaning ( |x(t)| \leq M ) for all ( t ) or ( |x[n]| \leq M ) for all ( n ), where ( M ) is some constant.
- System output: The output ( y(t) ) or ( y[n] ) of a system is given by the convolution of the input signal and the system’s impulse response:
- For continuous-time: ( y(t) = \int h(t – \tau) x(\tau) d\tau ).
- For discrete-time: ( y[n] = \sum_{k=-\infty}^{\infty} h[n-k] x[k] ).
- Effect of non-integrable or non-summable impulse response: Since ( h(t) ) or ( h[n] ) is not absolutely integrable or summable, the convolution operation can result in an unbounded output, even if the input is bounded.
- Example: If ( h(t) ) or ( h[n] ) grows large at certain time points (i.e., its absolute value does not decay quickly enough), then the integral or sum of the product of ( h(t) ) and ( x(t) ) or ( h[n] ) and ( x[n] ) can cause the output to become unbounded, even though the input is bounded.
Conclusion:
This demonstrates that the impulse response not being absolutely integrable or summable implies that there can be a bounded input leading to an unbounded output. This is a violation of BIBO stability, confirming that the conditions for integrability or summability are necessary for BIBO stability.