what is 20 log (-10) =?

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pot

Joined Nov 9, 2015
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Must be briefed constant value even start drawing , use this equation ===> 20 Log(-10)
 

shteii01

Joined Feb 19, 2010
4,644
Going on a limb here, but it strikes me as an imaginary number, similar to the square root of -1.
They are supposed to put frequency on the x-axis, and the absolute value of H(s) on the y-axis. Absolute value is Always positive. The correct formula is:
\(
Magnitude\quad in \quad dB=20*log_{10}(|H(s)|)
\)
 
Last edited:

WBahn

Joined Mar 31, 2012
25,902
what is 20 log (-10) =?
Your question lacks context. As a pure math question there are a couple of answers:

1) The normal logarithm function has the real numbers as both the domain and the range, hence the logarithm of a negative number is undefined since the base of normal logarithm functions are positive real numbers and there is no real exponent that you can apply to a positive real base and get a negative result.

2) The logarithm function can be extended to the complex numbers, in which the log of every non-zero argument is defined, though it is, in general, a multi-valued function.

But the look of the question implies that you are talking about voltage or current signals and converting them to decibels. In that case, you need to apply two functions, on for the magnitude and another for the phase. So you would have the magnitude as 20 log(10) and the phase as 180°.
 

WBahn

Joined Mar 31, 2012
25,902
Can you figure out a case where 10^x=-10?
Sure. In fact, there are infinitely many values of x for which 10^x yields -10. Here's one of them:

\(
x \; = \; 1 + j \frac{\pi}{\ln(10)}
\)

Let's check:

\(
10^x \; = \; 10^{\(1 + j \frac{\pi}{\ln(10)}\)}
\;
10^x \; = \; \( 10^1 \)\( 10^{\(j \frac{\pi}{\ln(10)}\)} \)
\)

Since

\(
10 \; = \; e^{\ln(10)}
\)

we have

\(
10^x \; = \; \( 10^1 \)\( {\(e^{\ln(10)}\)}^{\(j \frac{\pi}{\ln(10)}\)} \)
\;
10^x \; = \; 10e^{\(j \frac{\pi \ln(10)}{\ln(10)}\)}
\;
10^x \; = \; 10e^{j \pi}
\;
10^x \; = \; 10\(-1\)
\;
10^x \; = \; -10
\)
 
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