In a 4-bit ripple counter, if the period of the waveform at the last flip-flop is 64 microseconds, then the frequency of the ripple counter in kHz is ______.

The correct answer is 250.

A ripple counter (or asynchronous counter) consists of a series of flip-flops where the output of one flip-flop triggers the clock input of the next. For an n-bit ripple counter, each stage divides the frequency of the incoming signal by 2. Consequently, the overall frequency division factor for an n-bit counter is:

$$N=2^n$$

For a 4-bit ripple counter, the number of bits is n = 4. The division factor is:

$$N=2^4=16$$

The relationship between the frequency at the last flip-flop ($f_{\mathrm{out}}$) and the input clock frequency of the ripple counter ($f_{\mathrm{in}}$) is given by:

$$f_{\mathrm{out}}=\frac{f_{\mathrm{in}}}{16}$$

Since the time period is the reciprocal of frequency ($T=\frac{1}{f}$), the relationship between the time period of the waveform at the last flip-flop ($T_{\mathrm{out}}$) and the period of the input clock ($T_{\mathrm{in}}$) is:

$$T_{\mathrm{out}}=16\times T_{\mathrm{in}}$$

We are given that the period of the waveform at the last flip-flop is 64 microseconds ($T_{\mathrm{out}}=64\mathrm{\mu s}$). We can solve for the input period ($T_{\mathrm{in}}$):

$$T_{\mathrm{in}}=\frac{T_{\mathrm{out}}}{16}=\frac{64\mathrm{\mu s}}{16}=4\mathrm{\mu s}$$

Now, the frequency of the ripple counter (which is the frequency of the input clock, $f_{\mathrm{in}}$) is calculated as:

$$f_{\mathrm{in}}=\frac{1}{T_{\mathrm{in}}}=\frac{1}{4\times {10}^{-6}s}$$

Performing the division:

$$f_{\mathrm{in}}=250,000\mathrm{Hz}=250\mathrm{kHz}$$

Therefore, the frequency of the ripple counter is 250 kHz.