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-ⅈπfΔsinπfΔπf
P
f
f
Δ
f
Δ
f
(1)
Hf=11+ⅈ2πfRC
H
f
1
1
2
f
R
C
(2)
Thus, the output's Fourier transform equals
Yf=ⅇ-ⅈπfΔsinπfΔπf11+ⅈ2πfRC
Y
f
f
Δ
f
Δ
f
1
1
2
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-ⅈπfΔsinπfΔπf=ⅇ-ⅈπfΔⅇⅈπfΔ-ⅇ-ⅈπfΔⅈ2πf=1ⅈ2πf1-ⅇ-ⅈ2πfΔ
f
Δ
f
Δ
f
f
Δ
f
Δ
f
Δ
2
f
1
2
f
1
2
f
Δ
(4)
Consequently,
Yf=1ⅈ2πf1-ⅇ-ⅈπfΔ11+ⅈ2πfRC
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
1ⅈ2πf
1
2
f
means integration.
-
Multiplication by the complex exponential
ⅇ-ⅈ2πfΔ
2
f
Δ
means delay by
Δ
Δ
seconds in the time domain.
-
The term
1-ⅇ-ⅈ2πfΔ
1
2
f
Δ
means, in the time domain, subtract the time-delayed 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 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
f
(integration in the time domain) and the transfer function:
1ⅈ2πf11+ⅈ2πfRC↔1-ⅇ-tRCut
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-ⅈ2πfΔ
2
f
Δ
.
Because of the Fourier transform's linearity, we simply
subtract the results.
Yf
↔
1-ⅇ-tRCut-1-ⅇ-t-ΔRCut-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-ⅈ2πfΔ
1
2
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-ΔRCut-Δ
1
R
C
t
Δ
R
C
u
t
Δ
.
Subtracting from the undelayed signal yields
1RCⅇ-tRCut-1RCⅇ-t-ΔRCut-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
f
and
ⅇ-ⅈ2πfΔ
2
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
f
corresponded to a circuit that integrated, and
ⅇ-ⅈ2πfΔ
2
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 subproblems that are much easier to solve and then gluing the results together. 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
XX and the other with an
HH).
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 about constructing a transfer function from circuit elements and op-amps.
Another interesting notion arises from the commutative
property of multiplication (exploited in an
example above):
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
this Frequency Domain Problem. 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
(see this analog signal processing problem), we find a ready way
of realizing designs. We now understand why op-amp
implementations of transfer functions are so important.
"Electrical Engineering Digital Processing Systems in Braille."