First order lowpass filter, Exercises of Physics

First order lowpass filter. The first filter is a first order lowpass with cutoff frequency 1kHz, with transfer function. А(×) =.

Typology: Exercises

2022/2023

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Filtered Audio Demo
Max Kamenetsky
In this demo you'll listen to a 10 second segmentofmusic, alternating with various ltered
versions of it. You should try to relate what you hear to the frequency response, impulse
and step responses, and snapshots of the input and output signals.
First order lowpass lter
The rst lter is a rst order lowpass with cuto frequency 1kHz, with transfer function
H
(
s
)=
!
c
s
+
!
c
=
1
1+
s=!
c
;
where
!
c
=2
1000.
Note that it passes frequencies under around 500Hz or so, but attenuates high frequencies.
Since it attenuates high frequencies the ltered segment will sound a bit mued. (A higher
order lowpass lter, with a sharper cuto characteristic, would sound much more mued.)
The impulse response shows that this lter smooths out the input, giving a sort of averaging
over a few milliseconds. You can see that the ltered signal is a smoothed version of the
original signal.
102103104
−25
−20
−15
−10
−5
Frequency (Hz)
Magnitude (dB)
1−pole lowpass (Fc = 1kHz) Bode plot
102103104
−80
−70
−60
−50
−40
−30
−20
−10
Frequency (Hz)
Phase (deg)
1
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

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Filtered Audio Demo

Max Kamenetsky

In this demo you'll listen to a 10 second segment of music, alternating with various ltered versions of it. You should try to relate what you hear to the frequency resp onse, impulse and step resp onses, and snapshots of the input and output signals.

First order lowpass lter

The rst lter is a rst order lowpass with cuto frequency 1kHz, with transfer function

H (s) =

!c s + !c

1 + s=!c

where !c = 2   1000. Note that it passes frequencies under around 500Hz or so, but attenuates high frequencies. Since it attenuates high frequencies the ltered segment will sound a bit mued. (A higher order lowpass lter, with a sharp er cuto characteristic, would sound much more mued.) The impulse resp onse shows that this lter smo oths out the input, giving a sort of averaging over a few milliseconds. You can see that the ltered signal is a smo othed version of the original signal.

102 103 104

Frequency (Hz)

Magnitude (dB)

1−pole lowpass (Fc = 1kHz) Bode plot

102 103 104

Frequency (Hz)

Phase (deg)

0 1 2 3 4 5 6 7 8 x 10−

1000

2000

3000

4000

5000

6000

Time (s)

Amplitude

1−pole lowpass (Fc = 1kHz) impulse response

0 1 2 3 4 5 6 7 8 x 10−

0

Time (s)

Amplitude

1−pole lowpass (Fc = 1kHz) step response

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

Original signal

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

1−pole lowpass (Fc = 1kHz) filtered signal

−1 0 1 2 3 4 5 6 7 8 9 x 10−

0

2000

Time (s)

Amplitude

1−pole highpass (Fc = 1kHz) impulse response

−1 0 1 2 3 4 5 6 7 8 9 x 10−

0

1

Time (s)

Amplitude

1−pole highpass (Fc = 1kHz) step response

δ(t)

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

Original signal

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

1−pole highpass (Fc = 1kHz) filtered signal

Low Q bandpass lter

This is a second order bandpass lter, with center frequency 1 kHz and Q = 4. Transfer function: H (s) = K

s (Q=!c )s^2 + s + Q!c

where Q = 4, !c = 2   1000, and K = 6 : 7  16 : 5 dB is a scale factor to increase the volume level. Here frequencies b etween 500Hz and 2kHz or so are passed, and b oth low and high frequencies are attenuated. The acoustic e ect is something like listening through a tub e (which has resonances | you'll learn ab out that in EE141!). You can also hear when a note (or a harmonic) gets near 1kHz. The impulse resp onse is oscillatory, and you can see the e ect in the plot of the ltered signal. Here b oth the fast wiggles and the slow undulatations are attentuated.

102 103 104

0

10

Frequency (Hz)

Magnitude (dB)

First order bandpass (center = 1kHz, Q = 4) Bode plot

102 103 104

0

50

Frequency (Hz)

Phase (deg)

High Q bandpass lter

Second order bandpass lter with center frequency 1kHz and Q = 40. Transfer function:

H (s) = K

s (Q=!c )s^2 + s + Q!c

where Q = 40, !c = 2   1000, and K = 28  29 dB is a scale factor to increase the volume level. Here the frequencies near 1kH are strongly emphasized, which is quite annoying. The impulse resp onse is quite oscillatory, which you can also see in the plots of the ltered signal.

102 103 104

0

10

20

Frequency (Hz)

Magnitude (dB)

First order bandpass (center = 1kHz, Q = 40) Bode plot

102 103 104

0

50

Frequency (Hz)

Phase (deg)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.

0

2000

4000

Time (s)

Amplitude

First order bandpass (center = 1kHz, Q = 40) impulse response

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.

−0.

−0.

−0.

0

Time (s)

Amplitude

First order bandpass (center = 1kHz, Q = 40) step response

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

Original signal

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

First order bandpass (center = 1kHz, Q = 40) filtered signal

−1 0 1 2 3 4 5 6 7 8 9 x 10−

0

2000

4000

Time (s)

Amplitude

First order allpass impulse response

−1 0 1 2 3 4 5 6 7 8 9 x 10−

−0.

0

1

Time (s)

Amplitude

First order allpass step response

δ(t)

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

Original signal

0 0.5 1 1.5 2 2.5 3 3.5 4 4. x 10−

−0.

−0.

0

Time (s)

Amplitude

First order allpass filtered signal

Moving average lter

This lter is a 10ms moving average lter. Impulse resp onse:

h(t) =

K (1= 0 :1); 0  t < 0 : 01 0 ; t  0 : 01

The DC gain is K , which we take to b e K = 3 : 5  10 :9dB to increase the volume level. In this case the ltered signal output is exactly (K times the) average of the input signal over the last 10ms. Here the ltered signal sounds very mued; the high frequencies are strongly attenuated.

102 103 104

0

Frequency (Hz)

Magnitude (dB)

10 ms moving average Bode plot

Filter with echos

Impulse resp onse:

h(t) = Æ (t) + 0 : 75 Æ (t 0 :125) 0 : 1065 Æ (t 0 :1536) 0 : 4098 Æ (t 0 :1605) + 0 : 0308 Æ (t 0 :1788) + 0 : 0705 Æ (t 0 :1934) 0 : 2804 Æ (t 0 :2201) + 0 : 2906 Æ (t 0 :243) + 0 : 2898 Æ (t 0 :2567)

This represents 8 p erfect echos. The rst one arrives 125ms after the rst impulse, and the others come over the next 125ms or so.

102 103 104

0

5

10

Frequency (Hz)

Magnitude (dB)

Hall echos Bode plot

−0.5 0 0.1 0.2 0.3 0.4 0.5 0.

0

1

2

Time (s)

Amplitude

Hall echos impulse response

−0.5 0 0.1 0.2 0.3 0.4 0.5 0.

0

1

2

Time (s)

Amplitude

Hall echos step response

−0.2 0.227 0.2275 0.228 0.2285 0.229 0.2295 0.23 0.2305 0.231 0.

0

Time (s)

Amplitude

Original signal

−0.2 0.227 0.2275 0.228 0.2285 0.229 0.2295 0.23 0.2305 0.231 0.

0

Time (s)

Amplitude

Hall echos filtered signal