How do you find the forward kinematics of a 2-DOF serial robot using D-H homogeneous transformations?

In one line

Write the D-H table, form each link's transformation matrix, and multiply them; the last column of the result gives the end-effector position.

Explained simply

Each link transform tells you where the next joint is, seen from the previous one. Multiplying them is like adding up stage-by-stage directions to find the final destination from the start.

Technically, \( {}^0T_2 = A_1 A_2 \). For a planar RR arm both twists are zero, so each \( A_i \) is a rotation about z by \( \theta_i \) followed by a translation \( a_i \) along the new x-axis.

How it works

1. D-H table: link 1: \( (\theta_1, 0, L_1, 0) \); link 2: \( (\theta_2, 0, L_2, 0) \).
2. \( A_1 = \begin{bmatrix} c_1 & -s_1 & 0 & L_1c_1 \\ s_1 & c_1 & 0 & L_1s_1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \), with \( c_1 = \cos\theta_1 \), \( s_1 = \sin\theta_1 \); \( A_2 \) the same with index 2.
3. Multiply and simplify using \( c_1c_2 - s_1s_2 = c_{12} \), \( s_1c_2 + c_1s_2 = s_{12} \).
4. Read the position from the last column and the orientation from the 2×2 rotation block.
5. For an RP (polar) arm with a sliding second joint, the reach d becomes the variable: \( x = d\cos\theta \), \( y = d\sin\theta \).

Key formula

\( {}^0T_2 = \begin{bmatrix} c_{12} & -s_{12} & 0 & L_1c_1 + L_2c_{12} \\ s_{12} & c_{12} & 0 & L_1s_1 + L_2s_{12} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \), \( c_{12} = \cos(\theta_1 + \theta_2) \), \( s_{12} = \sin(\theta_1 + \theta_2) \). Lengths in m.

Quick example

\( L_1 = 0.4 \) m, \( L_2 = 0.3 \) m, \( \theta_1 = 45^\circ \), \( \theta_2 = 45^\circ \). \( c_1 = s_1 = 0.7071 \), \( c_{12} = 0 \), \( s_{12} = 1 \). Position: \( x = 0.4(0.7071) + 0 = 0.283 \) m, \( y = 0.283 + 0.3 = 0.583 \) m. The tool's x-axis points along +y of the base (rotated 90°).

Common mistakes

• Multiplying \( A_2A_1 \) instead of \( A_1A_2 \).
• Not simplifying to \( c_{12}, s_{12} \), which leaves long, error-prone expressions.
• Forgetting to check the answer with simple geometry of the arm.

How MU asks it

5 to 8-mark numerical: 'For the 2-DOF serial robot, assign frames, write the D-H table and find the end-effector position for given joint values'. Full marks: frame sketch, D-H table, \( A_1 \), \( A_2 \), product, numerical position. The syllabus limits this to 2-DOF serial robots.