- 2067 Asar (old course) · 8 marks
Forced response depends on input, natural does not
Short answer
Natural response form depends only on network elements (e.g., A e^(−2t) fixed by R/L). Forced response varies: 5 A (DC), 10e^(−t) (exponential), 4.47 sin(t−26.57°) (sinusoidal), proving it follows the input shape.
With a suitable example prove that the forced response of a network depends upon the nature of input excitation while the natural response never depends upon the input excitation.
Definitions. The natural (complementary) response is the solution of the homogeneous equation, with the source set to zero. Its form depends only on the network (R, L, C). The forced (particular) response is the part that follows the shape of the applied source.
Example: series R-L circuit
Take R = 2 Ω, L = 1 H with zero initial current. Equation: di/dt + 2i = v(t)/L = v(t).
Natural response (same for every input). Homogeneous equation di/dt + 2i = 0 gives i_n = A e^(−2t). It contains only the network constant R/L = 2.
Case 1: DC input v = 10 V
- Forced: i_f = 10/2 = 5 A (constant, same shape as the source)
- i = 5 + A e^(−2t); i(0) = 0 → A = −5
- i₁(t) = 5 − 5e^(−2t)
Case 2: exponential input v = 10 e^(−t) V
- Try i_f = K e^(−t): −K + 2K = 10 → K = 10, so i_f = 10e^(−t) (decays like the source)
- i = 10e^(−t) + A e^(−2t); i(0) = 0 → A = −10
- i₂(t) = 10e^(−t) − 10e^(−2t)
Case 3: sinusoidal input v = 10 sin t V. Forced response i_f = (10/√5) sin(t − 26.57°) = 4.47 sin(t − 26.57°). The natural term is again of the form A e^(−2t).
Comparison
| Input | Forced response | Natural response |
|---|---|---|
| 10 V (DC) | 5 A (constant) | A e^(−2t) |
| 10e^(−t) | 10e^(−t) | A e^(−2t) |
| 10 sin t | 4.47 sin(t − 26.57°) | A e^(−2t) |
The forced part changes with the source (constant, exponential, sinusoid). The natural part always has the exponent −R/L = −2 whatever the source, because it is the solution of the equation with the source removed. Only its constant A depends on the input, through the initial condition. This proves the statement.
Numerical check: Case 2 at t = 1 s gives i = 10(0.3679 − 0.1353) = 2.325 A, matching a numerical solution of di/dt + 2i = 10e^(−t).
✓ Answer checked by Claude Opus 5.5 (high effort)
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