For t > 1 ms, v ¼ 0:632eðt1Þ ¼ 1:72et , and i ¼ 1:72et (b) V0 ¼ 10 V, T ¼ 0:1 ms. For 0 < t < 0:1 ms, v ¼ 10ð1 et Þ; i ¼ 10et , and VT ¼ 10ð1 e0:1 Þ ¼ 0:95 V For t > 0:1 ms, v ¼ 0:95eðt0:1Þ ¼ 1:05et , and i ¼ 1:05et (c)
V0 ¼ 100 V, T ¼ 0:01 ms. For 0 < t < 0:01 ms, v ¼ 100ð1 et Þ 100t; i ¼ 100et 100ð1 tÞ, and VT ¼ 100ð1 e0:01 Þ ¼ 0:995 V For t > 0:01 ms, v ¼ 0:995eðt0:01Þ ¼ 1:01et and i ¼ 1:01et
As the input voltage pulse approaches an impulse, the capacitor voltage and current approach v ¼ et uðtÞ (V) and i ¼ ðtÞ et uðtÞ.
IMPULSE RESPONSE OF RC AND RL CIRCUITS
A narrow pulse can be modeled as an impulse with the area under the pulse indicating its strength. Impulse response is a useful tool in analysis and synthesis of circuits. It may be derived in several ways: take the limit of the response to a narrow pulse, to be called limit approach, as illustrated in Examples 7-11 and 7-12; take the derivative of the step response; solve the diﬀerential equation directly. The impulse response is often designated by hðtÞ. EXAMPLE 7.12 Find the limits of i and v of the circuit Fig. 7-17(a) for a voltage pulse of unit area as the pulse duration is decreased to zero. We use the pulse responses in (14) and (15) with V0 ¼ 1=T and ﬁnd their limits as T approaches zero. From (14c) we have lim VT ¼ lim ð1 eT=RC Þ=T ¼ 1=RC
From (15) we have: For t < 0;
hv ¼ 0
and hi ¼ 0 1 1 and hi ¼ ðtÞ 0 hv RC R 1 t=RC 1 hv ðtÞ ¼ and hi ðtÞ ¼ 2 et=RC e RC R C
For 0 < t < 0 ; For t > 0; Therefore, hv ðtÞ ¼
1 t=RC uðtÞ e RC
hi ðtÞ ¼
1 1 ðtÞ 2 et=RC uðtÞ R R C
EXAMPLE 7.13 Find the impulse responses of the RC circuit in Fig. 7-17(a) by taking the derivatives of its unit step responses. A unit impulse may be considered the derivative of a unit step. Based on the properties of linear diﬀerential equations with constant coeﬃcients, we can take the time derivative of the step response to ﬁnd the impulse response. The unit step responses of an RC circuit were found in (6) to be vðtÞ ¼ ð1 et=RC ÞuðtÞ
iðtÞ ¼ ð1=RÞet=RC uðtÞ
We ﬁnd the unit impulse responses by taking the derivatives of the step responses.
Published on May 10, 2013