The precision of plot filter design function is different than plot just coefficients of the same function of filter

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Johan Johan
Johan Johan 2020-1-27
评论： Star Strider 2020-1-27
I'm confused about plotting the design filter with the same precision ,for example i have this coefficients (i found it by using fir1(68,[.35,.65])):
bb= [0.0005
0
0.0010
0.0000
-0.0024
-0.0000
0.0020
-0.0000
0.0015
-0.0000
-0.0066
0.0000
0.0078
-0.0000
-0.0000
-0.0000
-0.0142
-0.0000
0.0220
-0.0000
-0.0093
0
-0.0232
-0.0000
0.0519
-0.0000
-0.0411
0.0000
-0.0305
-0.0000
0.1464
0.0000
-0.2551
0.0000
0.2995
0.0000
-0.2551
0.0000
0.1464
-0.0000
-0.0305
0.0000
-0.0411
-0.0000
0.0519
-0.0000
-0.0232
0
-0.0093
-0.0000
0.0220
-0.0000
-0.0142
-0.0000
-0.0000
-0.0000
0.0078
0.0000
-0.0066
-0.0000
0.0015
-0.0000
0.0020
-0.0000
-0.0024
0.0000
0.0010
0
0.0005];
%% plot
freqz(bb,1,512)
the design Fig01
if i use the same last funtion fir1 with same specification
b=fir1(68,[.35,.65])';
freqz(b,1,512)
the design in Fig02
Why the Fig1 don't like the Fig2 in stop band ,why don't find the same precision ?

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采纳的回答

Star Strider
Star Strider 2020-1-27
To reproduce the posted results, plot them as:
b = fir1(68,[.35,.65]);
bb = round(b, 4);
figure
freqz(bb,1,512);
figure
freqz(b',1,512)
In plotting ‘bb’, note that the coefficients have significantly reduced precision. Rounding (or truncating) the coefficients significantly affects the plotted filter frequency response.
  2 个评论
Johan Johan
Johan Johan 2020-1-27
Thank you, but i found this error
Error using round
Too many input arguments.
Does it the same idea with
b = fir1(68,[.35,.65]);
n=4;
bb = round(b*10^n)/10^n;
Star Strider
Star Strider 2020-1-27
As always, my pleasure!
Yes.
That should work.
I created my own version a few years ago:
roundn = @(x,n) round(x .* 10.^n)./10.^n; % Round ‘x’ To ‘n’ Digits, Emulates Latest ‘round’ Function
I do not now remember when the new round was introduced, although I believe it was R2016b, since many similar upgrades occurred them.

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