When we apply a periodic input to a linear, time-invariant system, the
output is periodic and has Fourier series coefficients equal to the product of the system's frequency response and the input's Fourier coefficients (Filtering Periodic Signals). The way we derived the spectrum of non-periodic signal from periodic ones makes
it clear that the same kind of result works when the input is not periodic: If
xt
x
t
serves as the input to a linear, time-invariant system having frequency response
Hf
H
f
, the spectrum of the output is
XfHf
X
f
H
f
.
Let's use this frequency-domain input-output relationship for linear, time-invariant 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−(iπfΔ)sinπfΔπf
P
f
π
f
Δ
π
f
Δ
π
f
(1)
Hf=11+i2πfRC
H
f
1
1
2
π
f
R
C
(2)
Thus, the output's Fourier transform equals
Yf=e−(iπfΔ)sinπfΔπf·11+i2πfRC
Y
f
π
f
Δ
π
f
Δ
π
f
·
1
1
2
π
f
R
C
(3)
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−(iπfΔ)sinπfΔπf=e−(iπfΔ)e∗iπfΔ−e−(iπfΔ)i2πf=1i2πf·(1−e−(i2πfΔ))
π
f
Δ
π
f
Δ
π
f
π
f
Δ
∗
π
f
Δ
π
f
Δ
2
π
f
1
2
π
f
·
1
2
π
f
Δ
(4)
Consequently,
Yf=1i2πf·(1−e−(iπfΔ))·11+i2πfRC
Y
f
1
2
π
f
·
1
π
f
Δ
·
1
1
2
π
f
R
C
(5)
The table of
Fourier transform properties suggests thinking about this expression as a
product of terms.
-
Multiplication by
1i2πf
1
2
π
f
means integration.
-
Multiplication by the complex exponential
e−(2πfΔ)
2
π
f
Δ
means delay by
Δ
Δ
seconds in the time domain.
-
The term
[1−e−(i2πfΔ)]
[
1
2
π
f
Δ
]
means, in the time domain, subtract from the original signal its time-delayed version.
-
The inverse transform of the frequency response is
1RCe−tRCut
1
R
C
t
R
C
u
t
.
We can translate each of these frequency-domain products into time-domain 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
1i2πf
1
2
π
f
(integration in the time domain) and the transfer function:
1i2πf·11+i2πfRC↔(1−e−tRC)ut
1
2
π
f
·
1
1
2
π
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−(i2πfΔ)
2
π
f
Δ
. Because of the Fourier transform's linearity, we simply subtract the results.
Yf
↔
(1−e−tRC)ut−(1−e−(t−Δ)RC)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 time-constant of the rising and falling portions of the output equal the product of the circuit's resistance and capacitance.
Derive the filter's output by considering
the terms in Equation 4 in the order given. Integrate last rather than first. You should get the same answer.
The inverse transform of the frequency response is
1RCe−tRCut
1
R
C
t
R
C
u
t
. Multiplying the frequency response by
[1−e−2πfΔ]
[
1
−
2
π
f
Δ
]
means subtract from the original signal its time-delayed version. Delaying the frequency response's time-domain version by
Δ
Δ
results in
1RCe−(t−Δ)RCut−Δ
1
R
C
t
Δ
R
C
u
t
Δ
. Subtracting from the undelayed signal yields
1RCe−tRCut−1RCe−(t−Δ)RCut−Δ
1
R
C
t
R
C
u
t
1
R
C
t
Δ
R
C
u
t
Δ
. Now we integrate this sum. Because the integral of a sum equals the sum of the component integrals (integration is linear), we can consider each separately. Because integration and signal-delay are linear, the integral of a delayed signal equals the delayed version of the integral. The integral is provided in the example.
In this example, we used the table extensively to find the inverse Fourier transform, relying mostly on what multiplication by certain factors, like
1i2πf
1
2
π
f
and
e−(2πfΔ)
2
π
f
Δ
, meant. We essentially treated multiplication by these factors as if they were transfer functions of some fictitious circuit. The transfer function
1i2πf
1
2
π
f
corresponded to a circuit that integrated, and
e−(2πfΔ)
2
π
f
Δ
to one that delayed. We even implicitly interpreted the circuit's transfer function as the input's spectrum! This approach to finding inverse transforms -- breaking down a complicated expression into products and sums of simple components -- is the engineer's way of breaking down the problem into several much simpler ones. Along the way we may make the system serve as the input, but in the rule
Yf=XfHf
Y
f
X
f
H
f
, which term is the input and which is the transfer function is merely a notational matter (we labeled one factor with an
X
X
and the other with an
H
H
).
The notion of a transfer function applies well beyond linear circuits. Although we
don't have all we need to demonstrate the result as yet, all linear, time-invariant
systems have a frequency-domain input-output relation given by the product of
the input's Fourier transform and the system's transfer function. Thus, linear
circuits are a special case of linear, time-invariant systems. As we tackle more
sophisticated problems in transmitting, manipulating, and receiving information,
we will assume linear systems having certain properties (transfer functions)
without worrying about what circuit has the desired property. At this point, you
may be concerned that this approach is glib, and rightly so. Later we'll show
that by involving software that we really don't need to be concerned in most
cases.
Another interesting notion arises from the commutative property of
multiplication (exploited in an example):
We can rather arbitrarily chose an
order in which to apply each product. Consider a cascade of two linear,
time-invariant systems. Because the Fourier transform of the first system's output is
Xf
H
1
f
X
f
H
1
f
and it serves as the second system's input, the cascade's output
spectrum is
Xf
H
1
f
H
2
f
X
f
H
1
f
H
2
f
. Because this
product also equals
Xf
H
2
f
H
1
f
X
f
H
2
f
H
1
f
, the cascade having the linear systems in the opposite order yields the same result.
Furthermore, the cascade acts like a single
linear system, having transfer function
H
1
f
H
2
f
H
1
f
H
2
f
. This result applies to other configurations of linear, time-invariant systems as well; see
Problem FIX ME. Engineers exploit this property by determining what transfer
function they want, then breaking it down into components arranged according to
standard configurations. Using the fact that op-amp circuits can be connected
in cascade with the transfer function equaling the product of its component's
transfer function (Problem FIX ME), we find a ready way of realizing designs.
We now understand why op-amp implementations of transfer functions are so
important.
