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

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 ?

Answer 1

[Star Strider]

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.

[Johan Johan]

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]

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.