# Clock Frequency and Duty Cycle

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?

• Your units are off. $1/10^{-9}s \neq 10^9s$, and you seem to be using $f$ instead of $T$ for the duty cycle. – Jasper Feb 17 '15 at 23:47
• Sorry, how is 1/10^-9s != 10^9s? I've retyped it in my calculator multiple times, and that seems to be the case... And presuming that the answer to part 1 is correct, T would then equal 10^15 instead of the negative I added in, right? – user215717 Feb 17 '15 at 23:55
• The units are off indeed. Frequency is measured in Hz, not is seconds. 1 ns period means 1 GHz frequency. – user58697 Feb 18 '15 at 0:42
• Okay, so the answer to number 1 is just 1GHz. Taking that, I would have for #2: D = (0.00035μs/0.001μs) * 100 = 35μs. Look better? – user215717 Feb 18 '15 at 0:49

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