Step 1: Recall definitions.
- Causality: A discrete-time LTI system is causal if its impulse response $h[n] = 0$ for all $n < 0$. Equivalently, its step response also vanishes for $n < 0$.
- Stability: An LTI system is stable (BIBO stable) if $\sum_{n=-\infty}^{\infty} |h[n]| < \infty$. This requires the impulse response to be absolutely summable.
Step 2: Evaluate option (A).
(A) "If an LTI system is causal, it is stable."
This is false, since causality and stability are independent properties.
Example: $h[n] = 1$ for $n \geq 0$. This is causal, but $\sum_{n=0}^{\infty} |1| = \infty$, so the system is unstable.
Step 3: Evaluate option (B).
(B) "A discrete time LTI system is causal iff step response $=0$ for $n<0$."
This is true. The step response is the cumulative sum of the impulse response:
\[
s[n] = \sum_{k=-\infty}^{n} h[k].
\]
If $h[k]=0$ for $k<0$, then $s[n]=0$ for $n<0$. Conversely, if $s[n]=0$ for $n<0$, then $h[k]=0$ for $k<0$.
Step 4: Evaluate option (C).
(C) "If $h[n]$ has finite duration, system is stable."
This is true, since $\sum |h[n]|$ is a finite sum if $h[n]$ is nonzero only for a finite number of $n$.
Step 5: Evaluate option (D).
(D) "If $0<|h[n]|<1$ for all $n$, then system is stable."
This is false. Even if $|h[n]|<1$, if it extends infinitely (like $h[n]=0.5$ for $n \geq 0$), then $\sum |h[n]| = \infty$, meaning unstable.
% Final Answer
\[
\boxed{\text{Correct statements: (B) and (C)}}
\]
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).
The relationship between two variables \( x \) and \( y \) is given by \( x + py + q = 0 \) and is shown in the figure. Find the values of \( p \) and \( q \). Note: The figure shown is representative.