Let's use this frequencydomain inputoutput relationship for
linear, timeinvariant systems to find a formula for the
RC
R
C
circuit's response to a pulse input. We have
expressions for the input's spectrum and the system's
frequency response.
Pf=e−(jπfΔ)sinπfΔπf
P
f
f
Δ
f
Δ
f
(1)
Hf=11+j2πfRC
H
f
1
1
2
f
R
C
(2)
Thus, the output's Fourier transform equals
You won't find this Fourier transform in our table, and the
required integral is difficult to evaluate as the expression
stands. This situation requires cleverness and an
understanding of the Fourier transform's properties. In
particular, recall Euler's relation for the sinusoidal term
and note the fact that multiplication by a complex exponential
in the frequency domain amounts to a time delay. Let's
momentarily make the expression for
Yf
Y
f
more complicated.
e−(jπfΔ)sinπfΔπf=e−(jπfΔ)ejπfΔ−e−(jπfΔ)j2πf=1j2πf(1−e−(j2πfΔ))
f
Δ
f
Δ
f
f
Δ
f
Δ
f
Δ
2
f
1
2
f
1
2
f
Δ
(4)
Consequently,
Yf=1j2πf(1−e−(jπfΔ))11+j2πfRC
Y
f
1
2
f
1
f
Δ
1
1
2
f
R
C
(5)
The table of
Fourier transform properties suggests
thinking about this expression as a
product of terms.

Multiplication by
1j2πf
1
2
f
means integration.

Multiplication by the complex exponential
e−(j2πfΔ)
2
f
Δ
means delay by
Δ
Δ
seconds in the time domain.

The term
1−e−(j2πfΔ)
1
2
f
Δ
means, in the time domain, subtract the timedelayed signal from its original.

The inverse transform of the frequency response is
1RCe−tRCut
1
R
C
t
R
C
u
t
.
We can translate each of these frequencydomain products into
timedomain operations
in any order we
like because the order in which multiplications
occur doesn't affect the result. Let's start with the product
of
1j2πf
1
2
f
(integration in the time domain) and the transfer function:
1j2πf11+j2πfRC↔(1−e−tRC)ut
1
2
f
1
1
2
f
R
C
↔
1
t
R
C
u
t
(6)
The middle term in the expression for
Yf
Y
f
consists of the difference of two terms: the constant
1
1
and the complex exponential
e−(j2πfΔ)
2
f
Δ
.
Because of the Fourier transform's linearity, we simply
subtract the results.
Yf
↔
(1−e−tRC)ut−(1−e−t−ΔRC)ut−Δ
Y
f
↔
1
t
R
C
u
t
1
t
Δ
R
C
u
t
Δ
(7)
Note that in delaying the signal how we carefully included the
unit step. The second term in this result does not begin until
t=Δ
t
Δ
. Thus, the waveforms shown in the
Filtering
Periodic Signals example mentioned above are
exponentials. We say that the
time constant of
an exponentially decaying signal equals the time it takes to
decrease by
1e
1
of its original value. Thus, the timeconstant of the rising
and falling portions of the output equal the product of the
circuit's resistance and capacitance.
"Electrical Engineering Digital Processing Systems in Braille."