ASINH

Returns the inverse hyperbolic sine of a number.

Syntax:

ASINH(number)

returns the inverse hyperbolic sine of number, in other words the number whose hyperbolic sine is number.

Example:

ASINH(-90)

returns approximately -5.1929877.

ASINH(SINH(4))

returns 4.

Application:

Analyzing the Current in a Long Transmission Line

Imagine a very long electrical transmission line, like those spanning across continents, with a sinusoidal voltage source at one end. The current distribution along such a line doesn't follow a simple linear relationship. Instead, it's described by hyperbolic functions due to the line's distributed resistance, inductance, capacitance, and conductance.

The current at any point 'x' along the line can be modeled by an equation that includes hyperbolic sine and cosine terms. Let's simplify this to focus on the role of ASINH.

Suppose we have a formula that relates the current at a certain point 'x' to the current at the receiving end of the line, which involves the hyperbolic sine function.

$I (x) = I_{R} cosh (γ x) + \frac{V _{R}}{Z _{c}} sinh (γ x)$

Here:

I(x) is the current at a distance 'x' from the receiving end.
IR and VR are the current and voltage at the receiving end, respectively.
Zc is the characteristic impedance of the line.
γ is the propagation constant of the line.

Now, let's say we want to solve for the distance 'x' at which the current reaches a specific value, but we only know the relationship between the current and the hyperbolic sine of some value. For a simplified case, let's consider a scenario where we can isolate the hyperbolic sine term.

Let's assume we have a simplified model where the current's behavior is directly related to the hyperbolic sine of the distance from the source. The relationship is:

$I = I_{ma x} sinh (αx)$

where:

I is the current at a certain point.
Imax is a constant representing the maximum possible current under certain conditions.
α is a constant related to the line's properties.
x is the distance from the source.

Our goal is to find the distance 'x' for a given current 'I'. To do this, we need to solve for 'x':

$\frac{I}{I _{ma x}} = sinh (αx)$

To isolate 'x', we must use the inverse hyperbolic sine function:

$ASINH (\frac{I}{I _{ma x}}) = αx$

$x = \frac{1}{α} ASINH (\frac{I}{I _{ma x}})$

Let's use some hypothetical values to build a table. Suppose Imax=100 A and α=0.05 km−1. We want to find the distance 'x' for different current values.

	Current (I, in Amperes)	I/Imax	ASINH(I/Imax)	Distance (x = 1/0.05 ASINH(I/Imax), in km)
	A	B	C	D
1	10	0.1	0.099834079	1.996681578
2	50	0.5	0.481211825	9.624236501
3	100	1	0.881373587	17.62747174
4	150	1.5	1.194763217	23.895264346
5	200	2	1.443635475	28.872709504

Explanation of the Table:

Current (I): These are the specific current values we are measuring at different points along the transmission line.
I/Imax: This is the ratio of the measured current to the maximum current, which is the input to the ASINH function.
ASINH(I/Imax): This column shows the result of applying the inverse hyperbolic sine function to the ratio.
Distance (x): This is the final calculated distance from the source at which the corresponding current value is observed. As the current increases, the required distance along the transmission line also increases.

This example demonstrates how the ASINH function is used to "undo" a hyperbolic sine relationship, allowing engineers to solve for a physical parameter (distance) when the measurement (current) is known, which is a common task in the analysis of distributed parameter systems like transmission lines.