Making synthetic formants

We now analyze the spectrum of the periodic signal generated in Figure 2 in the special case where the table $p(t)$ (still regarded as holding a continuous function of the parameter $t$ ) contains a sinusoid, either the fundamental or a harmonic:

$$p(t) = {e^{2 \pi h t i}}$$

where the integer $h$ is the number of the harmonic. Since the output is a linear function of the waveform, we can then infer the spectrum for an arbitrary choice of $p(t)$ .

Figure 3: Spectrum of the output of Figure 2: a. $hS = 3$ , $T=1$ ; b. $hS = 3.5$ , $T=1$ ; c. $hS = 3.75$ , $T=2$ .

For simplicity of analysis we will suppose the fundamental is of the form $\omega = 2\pi/N$ so that the resulting period $N$ is an integer; this only amounts to a possible sample rate adjustment which does not affect the end result. For $-N \le n < N$ the output of the diagram of Figure 1 (with period $2N$ ) is:

$$x[n] = w(nT/2N) p(nS/N)$$

This may be expressed as a Fourier series:

$$x[n] = \sum_{k=0}^{N-1} {b_k} {e^{2 \pi k t i/ (2N)}}$$

where the partial amplitudes $b_k$ are real-valued (phase zero at $n=0$ ) and equal to:

$${b_k} = {1 \over T}W({{k - 2hS}\over{T}})$$

$$W(k) = {1\over 4} \mathrm{sinx}(k-1) + {1\over 2} \mathrm{sinx}(k) + {1\over 4} \mathrm{sinx}(k+1)$$

$$\mathrm{sinx}(k) = \sin(2\pi k)/(2 \pi k) \; \;$$   $$\mbox{(or $1$\ when $k=0$)}$$

Here $W(k)$ is the (suitably normalized) Fourier transform of the Hann window function. Overlap-adding as in Figure 2 extracts the even-numbered partials:

$$y[n] = x[n] + x[n+N]$$

which gives:

$$y[n] = \sum_{k=0}^{N-1} {c_k} {e^{2 \pi k t i/ N}}$$

$${c_k} = {2 \over T}W({{2(k - hS)}\over{T}})$$

Figure 3 shows the result for three $(hS,T)$ pairs. The spectrum has one formant. Changing the parameter $S$ has the effect of rescaling the spectral envelope along the frequency axis. The parameter $T$ controls bandwidth.

When the minimum bandwidth is selected by setting $T=1$ , as in parts (a) and (b) of the figure, the result is a mixture of consecutive partials; if the center frequency lies directly on a partial, only the one corresponding partial is present in the result. This is the same result as that of setting the bandwidth to zero in the PAF generator.