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I have two signals. One has a frequency of $10$ Hz, the other has a frequency of $5$ Hz.

Or, in other words, one has a period of $\dfrac 1{10}$ seconds, and the other has a $\dfrac 15$ seconds.

The difference in their frequencies is $10 Hz - 5 Hz = 5 Hz$.

The difference in their periods is $\dfrac 1{10} - \dfrac 15 = -0.1$ seconds.

But $\dfrac 15 Hz$ = $2$ seconds $!= -0.1$ seconds.

Where am I going wrong here?

rhombidodecahedron
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    $$1/a - 1/b \ne 1 / (a - b)$$ so why would you expect them to be the same? –  Mar 01 '18 at 18:42
  • @user296602 Yes I see that. But shouldn't the differences be somehow relatable, since they are expressed in complementary units (seconds and Hz)? – rhombidodecahedron Mar 01 '18 at 18:43
  • The are related, through the correct equality $1/a - 1/b = (b - a) / (ab)$, which has the same units as $1 / (a - b)$. –  Mar 01 '18 at 18:44
  • So to put things differently, if I know (only) the difference in frequency, I don't necessarily know the difference in period? – rhombidodecahedron Mar 01 '18 at 18:49
  • Yes. Think about specific examples: What is the difference in period between 1 Hz and 2 Hz? What is the difference in period between 100 Hz and 101 Hz? Are they the same? –  Mar 01 '18 at 18:50
  • The difference in the frequencies or periods really has no actual relevance to anything - what matters is the ratio. – David C. Ullrich Mar 01 '18 at 18:56

2 Answers2

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As observed by MalayTheDynamo note that since

$$f=\frac1T \implies f_2-f_1=\frac1{T_2}-\frac1{T_1}=\frac{T_1-T_2}{T_2T_1}$$

thus

$$10 Hz-5Hz =5Hz = \frac{0.20-0.10}{0.20\cdot 0.10}=\frac{0.10}{0.02}= 5 Hz$$

user
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You're saying that $\frac1a+\frac1b=\frac1{a+b}$. This is true.

What you want is to add, then multiply by product.

$$\frac1a+\frac1b=\frac{a+b}{ab}$$

You're making an elementary mistake, and this is usual. Don't worry, you'll get better rapidly with time.

DynamoBlaze
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