Plot signal of DFT without using FFT function

I looked at so many blogs until I came here and still can't figure it out.

I have a signal given as:

The DFT of that signal is:

The analytical solution for fk is:

Where fk = 100 when k = 67 and 0 otherwise.

Here's what I have:

clc;clear;
N = 715;
k = 0:N-1;
n = k;

fj_func = @(N,n) (100*exp(2i.*pi.*67/N.*k)).*(n>=0 & n<=N-1);
fj = fj_func(N,n);
fk_func = @(N,k) (1/N).*fj.*(exp(-2i.*pi.*k.*n./N)).*(k>=0 & k<=N-1);
fk = fk_func(N,k);

stem(k,abs(fk)/N)

When I use the analytical solution above, I get:

clc;clear;
N = 715;
k = 0:N-1;
n = k;

fk_func = @(k) (100/N).*((1-exp(2i.*pi.*n.*(67-k)))./(1-exp(2i.*pi.*n.*(67-k)./N)));
fk = fk_func(k);

stem(k,abs(fk)/N)

I used the fft function for fj, and it works. My value is at

clc;clear;
N = 715;
k = 0:N-1;
n = k;

fj_func = @(N,k) (100*exp(2i.*pi.*67/N.*k)).*(k>=0 & k<=N-1);
fj = fj_func(N,k);

ft = fft(fj);
stem(k,abs(ft)/N)

I'm not sure what I'm doing wrong. Any ideas would be greatly appreciated. Thanks

Answer 1

[Paul]

The first approach to compute DFT directly using the definition can't be correct because there is no summation over n. Try it this way. First off there really isn't a need to define a function to compute fn.

N = 715;
n = 0:(N-1);
fn = 100*exp(2i*pi*67*n/N);

Now define a function to compute the DFT for a given value of k

dftfn = @(k,fn,n,N) (sum(fn .* exp(-1i*2*pi*k*n/N))/N);

Now test for a couple of values of k

dftfn(5,fn,n,N)

ans = -1.9339e-14 - 3.3649e-14i

dftfn(67,fn,n,N)

ans = 1.0000e+02 - 1.4181e-16i

The next step would be to write a loop to evaluate dftfn at the values of k of interest.

Note: there are other alternatives, but I think this is the clearest.

For the second approach, I don't think the closed form expression is correct. There shoudn't be an n in any term.

fk_func = @(k,N) (100/N).*((1-exp(2i.*pi.*(67-k)))./(1-exp(2i.*pi.*(67-k)./N)));
k = 0:(N-1);
fk = fk_func(k,N);
stem(k,abs(fk))  % no need to divide by N, that's already in fk_func

As expected all the reults are tiny, but what happened at k = 67?

stem(k,abs(fk));
xlim([60 70])

We see that there is a gap, because the formula at k = 67 yields NaN because it evaluates to 0/0.

fk_func(67,N)

ans = NaN

[Nicholas Saint]

I completely forgot about the summation which was staring me at the face. Thank you so much!