Clock Frequency and Duty Cycle

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A clock has a 1ns clock period with rise and fall time as 0.05ns. The clock signal stays at exact Boolean state 1 for 0.35ns and at state 0 for 0.55ns. The memory used in the design takes 2 clock cycle time to compute a write and 1 clock cycle to compute a read operation.

  1. What is the frequency of this clock? My attempt: $T = 1/f \Rightarrow f = 1/T = 1/1ns = 1/10^{-9}s = 10^9s = 10^{15}μs$

  2. What is the duty cycle of this clock? My attempt: $D = t_hh/T * 100 = (0.35ns/10^{-15}μs) * 100 = 0.00035μs/10^{-15}μs = 3.5^{13}μs$

Could someone please kindly confirm whether I did this correctly or not?

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I've retyped it in my calculator multiple times, and that seems to be the case.

The calculator sure is right about the numbers, but you didn't enter any units, right?

Regarding the frequency, you are right for this part $$T=1/f \Rightarrow f=1/T=1/1\text{ns}=1/10^{−9}\text{s}$$ Note that the last term is to be read as $$\frac{1}{10^{-9}\text{s}} = 10^9 \frac{1}{\text{s}} = 10^9 \text{Hz} = 1 \text{GHz}$$ and not as $$\frac{1}{10^{-9}}\text{s}$$

This hint usually applies to physics, and basically every time you deal with units: if the unit of the result is wrong, the result itself is wrong. If you are asked for a frequency (number of times something occurs in a given time span) and you come up with a time span, that's wrong.

The same thing applies to the second question: if you divide a time by a time, you get a dimensionless number. $$D = t_h/T \cdot 100\% = \frac{0.35 \cdot 10^{-9}\text{s}}{1 \cdot 10^{-9}\text{s}}\cdot 100\% = 35\%$$