IEEE Solid-States Circuits Magazine - Summer 2020 - 11

; H (z) ; does not fall to zero for any
sinusoidal input because a delayed
and scaled copy of the input cannot yield complete cancellation unless a = 1. For small values of a, the
LPF exhibits little selectivity [Figure 7(c)].
Let us further extend our impulse
broadening notion and add more
delayed replicas of the input. Figure 8(a) illustrates an example, for
which we have
	

sponse of a CT HPF, sketched in Figure 9(a). Note that y(t) must decay to zero
because the circuit does not pass dc.
Taking the derivative, we obtain the
impulse response, Figure 9(b), observing that the output contains a positive
impulse and a negative broadened tail.
This points to the DT counterpart in
Figure 9(c) and the realization, in Figure 9(d), where H (z) = Y/X = 1 - z -1.
Sketched in Figure 9(e), the filter's frequency response begins from zero at
f = 0, where z -1 " 1, and reaches 2 at
f = fCK /2, where z -1 " - 1.

H (z) = 1 + z -1 + z -2.(15)

We expect that the greater broadening of the input impulse leads to
a narrower bandwidth for the filter. Indeed, from the step response
shown in Figure 8(b), we observe a
more gradual transition. Equation
(15) exemplifies a simple finite impulse response filter. A more general form is

δ [k ]

0

0

0

k

x

u [k ]

y

2

3

0

k

δ (t )

x
0

t

0

4

k

h (t )

y

0

t

t

(b)

y

h [k ]

+

x [k ]

1

y [k ]
-

0

k

3

HPF

(a)

0

1

FIGURE 8: An LPF with greater impulse broadening: the (a) impulse response and
(b) step response.

y

HPF

2

k

(b)

t

x

2

1

LPF
1

δ [k ]

1

(a)

2

The developments in the previous
section can be repeated for high-pass
filters (HPFs) as well. In this case, it
is simpler to begin with the step re-

HPF

h [k ]

3

High-Pass Filters

x

y
LPF

	 H (z) = 1 + a 1 z -1 + a 2 z -2 + g.(16)

u (t )

x

Figure 9(d) implies that subtracting a delayed replica of a signal
from itself performs high-pass filtering. This agrees with our previous results: a slowly varying signal
and its delayed copy nearly cancel
each other upon subtraction.
Our first-order HPF merits one
more remark. We plot the step response as shown in Figure 10, recognizing that the two signals cancel
each other except at k = 0. Thus, the
step response is simply an impulse,
revealing that 1 - z -1 represents a

k

z -1

(c)

(d)
H (z )
2

0

fCK
(e)

f

2

FIGURE 9: The (a) step response of a CT HPF, (b) resulting impulse response, (c) DT HPF impulse response, (d) DT HPF realization, and
(e) DT HPF frequency response.

	

	 IEEE SOLID-STATE CIRCUITS MAGAZINE	

SU M M E R 2 0 2 0	

11



IEEE Solid-States Circuits Magazine - Summer 2020

Table of Contents for the Digital Edition of IEEE Solid-States Circuits Magazine - Summer 2020

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